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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 201. |
The plot of `sqrt(v)` vs Z isA. Straight lineB. Exponential curveC. HyperbolicD. Curve with -ve slope |
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Answer» Correct Answer - A |
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| 202. |
Chemical similarity between Be and Al is due toA. Diagonal relationshipB. Both belong to same periodC. Similar outer electronic configurationD. Inert pair effect |
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Answer» Correct Answer - A |
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| 203. |
The longest and shortest periods areA. 1 & 6B. 2 & 6C. 6 & 1D. 1 & 7 |
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Answer» Correct Answer - C |
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| 204. |
Pair of ions with polarising powerA. `Li^(+), Mg^(2+)`B. `Li^(+), Na^(+)`C. `Mg^(2+), Ca^(2+)`D. `Mg^(2+), K^(+)` |
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Answer» Correct Answer - A |
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| 205. |
The atomic numbers of Lanthanides are fromA. 58 to 71B. 90 to 103C. 21 to 30D. 39 to 48 |
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Answer» Correct Answer - A |
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| 206. |
A member of LanthanideA. CaesiumB. LanthanumC. NiobiumD. Lutitium |
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Answer» Correct Answer - D |
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| 207. |
A member of LanthanideA. CaesiumB. LanthanumC. NiobiumD. Luticium |
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Answer» Correct Answer - D |
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| 208. |
Which is wrong about transition metals?A. They are diamagneticB. they are paramagneticC. They form complexesD. They show variable oxidation states |
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Answer» Correct Answer - A |
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| 209. |
Which of the following statements is wrong for the transition elements?A. Transition elements are placed from 3rd to 6th periodB. Last electron enters in `(n-d)d` subshellC. Exhibits variable valencyD. General electronic configuration is `(n-1) d^(1-10) ns^(1-2)`. |
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Answer» Correct Answer - A Transition elements elements 4th period 7-th period |
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| 210. |
Numerous forms of the periodic table have been revised from time to time. A modern verion, which is most convenient and widely used is the long or extended from the periodic table. The aufbau principle (electrons are filled in the progressive order of their increasing energy i.e., by n + l rule) and the electronic configuration of atom provides a theoretical foundation for the periodic classification. The horizontal rows are called periods. There are altogether seven periods. The first period consists of 2 elements. The subsequent periods consist of 8, 8, 18 and 32 elements respectively. The seventh maximum of 32 elements. Elements having similar outer electronic configurations in their atoms are grouped in vertical columns. These are referred to as groups or famillies. According to the recommendation of IUPAC, the groups are numbered 1 to 18 replacing the older notation of groups 0, IA, IIA ..... VIIA, VIII, IB, ...... VII B. Each successive period in the periodic table is associated with the filling up of next higher principal energy level following aufbua sequence. The number of elements in each period is twice the number of atomic orbitals avaliable in the energy level that is being filled. All the elements are classified into four blocks, i.e., s-block, p-block, d-block and f-block depending on the type of atomic orbitals that are being filled with the last electron of the element. If aufbau rule is not followed, Ca-20 will be placed in........block.A. sB. pC. dD. f |
| Answer» Correct Answer - C | |
| 211. |
Numerous forms of the periodic table have been revised from time to time. A modern verion, which is most convenient and widely used is the long or extended from the periodic table. The aufbau principle (electrons are filled in the progressive order of their increasing energy i.e., by n + l rule) and the electronic configuration of atom provides a theoretical foundation for the periodic classification. The horizontal rows are called periods. There are altogether seven periods. The first period consists of 2 elements. The subsequent periods consist of 8, 8, 18 and 32 elements respectively. The seventh maximum of 32 elements. Elements having similar outer electronic configurations in their atoms are grouped in vertical columns. These are referred to as groups or famillies. According to the recommendation of IUPAC, the groups are numbered 1 to 18 replacing the older notation of groups 0, IA, IIA ..... VIIA, VIII, IB, ...... VII B. Each successive period in the periodic table is associated with the filling up of next higher principal energy level following aufbua sequence. The number of elements in each period is twice the number of atomic orbitals avaliable in the energy level that is being filled. All the elements are classified into four blocks, i.e., s-block, p-block, d-block and f-block depending on the type of atomic orbitals that are being filled with the last electron of the element. Elements a,b,c,d and e have the following electronic configurations. (a) `1s^(2),2s^(2)2p^(1)` (b) `1s^(2),2s^(2)2p^)(6),3s^(2)3p^(1)` (c) `1s^(2),2s^(2)2p^(6),3s^(2)3p^(1)` (d) `1s^(2),2s^(2)2p^(6),3s^(2)3p^(5)` (e) `1s^(2), 2s^(2)2p^(6),3s^(2)3p^(5)` Which among these will belong to the same group in periodic table?A. a and cB. a and bC. a and bD. d and e |
| Answer» Correct Answer - B | |
| 212. |
Numerous forms of the periodic table have been revised from time to time. A modern verion, which is most convenient and widely used is the long or extended from the periodic table. The aufbau principle (electrons are filled in the progressive order of their increasing energy i.e., by n + l rule) and the electronic configuration of atom provides a theoretical foundation for the periodic classification. The horizontal rows are called periods. There are altogether seven periods. The first period consists of 2 elements. The subsequent periods consist of 8, 8, 18 and 32 elements respectively. The seventh maximum of 32 elements. Elements having similar outer electronic configurations in their atoms are grouped in vertical columns. These are referred to as groups or famillies. According to the recommendation of IUPAC, the groups are numbered 1 to 18 replacing the older notation of groups 0, IA, IIA ..... VIIA, VIII, IB, ...... VII B. Each successive period in the periodic table is associated with the filling up of next higher principal energy level following aufbua sequence. The number of elements in each period is twice the number of atomic orbitals avaliable in the energy level that is being filled. All the elements are classified into four blocks, i.e., s-block, p-block, d-block and f-block depending on the type of atomic orbitals that are being filled with the last electron of the element. The element with atomic number 56 is likely to have the same outer shell configuration as the element with atomic number.A. 12B. 18C. 14D. 20 |
| Answer» Correct Answer - A::D | |
| 213. |
Numerous forms of the periodic table have been revised from time to time. A modern verion, which is most convenient and widely used is the long or extended from the periodic table. The aufbau principle (electrons are filled in the progressive order of their increasing energy i.e., by n + l rule) and the electronic configuration of atom provides a theoretical foundation for the periodic classification. The horizontal rows are called periods. There are altogether seven periods. The first period consists of 2 elements. The subsequent periods consist of 8, 8, 18 and 32 elements respectively. The seventh maximum of 32 elements. Elements having similar outer electronic configurations in their atoms are grouped in vertical columns. These are referred to as groups or famillies. According to the recommendation of IUPAC, the groups are numbered 1 to 18 replacing the older notation of groups 0, IA, IIA ..... VIIA, VIII, IB, ...... VII B. Each successive period in the periodic table is associated with the filling up of next higher principal energy level following aufbua sequence. The number of elements in each period is twice the number of atomic orbitals avaliable in the energy level that is being filled. All the elements are classified into four blocks, i.e., s-block, p-block, d-block and f-block depending on the type of atomic orbitals that are being filled with the last electron of the element. What is the position of the element in the periodic table satisfying the electronic configuration `(n1)d^(1)ns^(2)` for `n=4`?A. 3rd period and 3rd groupB. 4th period and 4th groupC. 3rd period and 2nd groupD. 4th period and 3rd group |
| Answer» Correct Answer - D | |
| 214. |
Which one of the following statements related to the modern periodic table is incorrect?A. The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-subshellB. The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshellC. Each block contains a number of columns equal to the number of electrons that can occupy that subshell.D. The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration. |
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Answer» Correct Answer - B d-block has 10 coloumns |
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| 215. |
In 1931, Pauling defined the electronegativity of an atom as the tendency of the atom to attract electrons to itself when combined in a compound. The implication is that when a covalent bond is formed, the electrons used for bonding need not be shared equally by both atoms. If the bonding electrons spend more time around one atom, that atom will have a `del-`charge, and consequently the other atom will have a `del+` charge. In the extreme case, where the bonding electrons are round one atom all of the time, the bond is ionic. Pauling and others have attempted to relate the electronegativity difference between two atoms to the amount of ionic character in the bond between them. Mulliken In 1934, Mulliken suggested an alternative approach to electronegativity based on the ionization energy and electron affinity of an atom. Consider two atoms A and B. If an electron is transferred from A to B, forming ions `A^(+)` and `B^(-)`, then the energy change is the ionization energy of atom `A(I_(A))` minus the electron affinity of atom `B(E_(B))`, that is `I_(A)-E_(B)`. Alternatively, if the electron was transferred the other way to give `B^(+)` and `A^(-)` ions, then the energy changed would be `I_(B) - I_(A)`. If `A^(+)` and `B^(-)` are actually formed, then this process require less energy , and `(I_(A) - E_(B)) lt (I_(B) - E_(A))` Rerranging `(I_(A)+E_(A)) lt (I_(A)-E_(B))` Now with respect to electronegativity for the same change, `E.N_(A) lt E.N_(B)`. Thus Mulliken suggested that electronegativity is proportional to `I.E + E.A` and could be regarded as the average of the ionization energy and the electron affinity of an atom. Electronegativity `=((I+E))/(2)` Mulliken used I and E values measured in electron volts, and the values were about 2.8 times alrger than the Pauling values. It is to be noted that `I.E`. and `E.A` values are defined for singular gaseous atoms. For the reaction `A(g) +e^(-) rarr A^(+)(g) Delta H_(r) = DeltaH_(eg)`. The enthalpy change will be equal to (magnitude wise):A. `DeltaH_(eg)` of `A^(-)(g)`B. `DeltaH_(I.E)` of `A(g)`C. `DeltaH_(E.N)` of `A(g)`D. `DeltaH_(I.E)` of `A^(-)(g)` |
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Answer» Correct Answer - D `A^(-)(g) rarr a(g)+e^(-) DeltaH_(r) = DeltaH_(I.E)` of `A^(-)(g)` |
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| 216. |
In 1931, Pauling defined the electronegativity of an atom as the tendency of the atom to attract electrons to itself when combined in a compound. The implication is that when a covalent bond is formed, the electrons used for bonding need not be shared equally by both atoms. If the bonding electrons spend more time around one atom, that atom will have a `del-`charge, and consequently the other atom will have a `del+` charge. In the extreme case, where the bonding electrons are round one atom all of the time, the bond is ionic. Pauling and others have attempted to relate the electronegativity difference between two atoms to the amount of ionic character in the bond between them. Mulliken In 1934, Mulliken suggested an alternative approach to electronegativity based on the ionization energy and electron affinity of an atom. Consider two atoms A and B. If an electron is transferred from A to B, forming ions `A^(+)` and `B^(-)`, then the energy change is the ionization energy of atom `A(I_(A))` minus the electron affinity of atom `B(E_(B))`, that is `I_(A)-E_(B)`. Alternatively, if the electron was transferred the other way to give `B^(+)` and `A^(-)` ions, then the energy changed would be `I_(B) - I_(A)`. If `A^(+)` and `B^(-)` are actually formed, then this process require less energy , and `(I_(A) - E_(B)) lt (I_(B) - E_(A))` Rerranging `(I_(A)+E_(A)) lt (I_(A)-E_(B))` Now with respect to electronegativity for the same change, `E.N_(A) lt E.N_(B)`. Thus Mulliken suggested that electronegativity is proportional to `I.E + E.A` and could be regarded as the average of the ionization energy and the electron affinity of an atom. Electronegativity `=((I+E))/(2)` Mulliken used I and E values measured in electron volts, and the values were about 2.8 times alrger than the Pauling values. It is to be noted that `I.E`. and `E.A` values are defined for singular gaseous atoms. We have X atoms of A. if all the atoms gain one electron each the energy released is `aeV`. If Y atoms of A lose one electron each, then the energy absorbed is beV. Then the electronegativity of A on the Mulliken scale will be:A. `(a+b)/(2)`B. `(1)/(2)((a)/(X)+(b)/(Y))`C. `(1)/(2)(a+b)xx N_(a)`D. `(1)/(2)((a)/(X)+(b)/(Y))xx N_(a)` |
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Answer» Correct Answer - B For A, electron affinity per atom `= (a)/(X) ev` For A, ionisation energy per atom `=(b)/(Y) ev` `E.N.` (Mulliken) `=(1)/(2) ((a)/(X)+(b)/(Y))` |
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| 217. |
The energy required to pull the most loosely bound electrons from an atom is known as ionization potential. It is expressed in electron volts. The value of ionization potential depends on three factors :(i) the charge on the nucleus (ii) the atomic radius and (iii) the screening effect of inner electron shells. Incorrect order of ionisation enegy is:A. `Pb (I.E.) gt Sn (I.E.)`B. `Na^(+) (I.E.) gt Mg^(+) (I.E.)`C. `Li^(+) (I.E.) lt O^(+) (I.E.)`D. `Be^(+) (I.E.) lt C^(+) (I.E.)` |
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Answer» Correct Answer - C |
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| 218. |
The minimum amount of energy which is required to remove an outermost electron from any isolated neutral gaseous atom is known as first ionisation energy. These are the following factors which effect ionisation energy. (i) Ionisation energy `prop (1)/("Principal quantum number")` (ii) Ionisation energy `prop Z_(eff)` (iii) If orbitals are fully filled or half filled so stability will be more and ionisation energy will be high. (iv) If penultimate electron will effectively shield the nucles, ionisation energy will be less and vice versa. Choose the correct order of Ist ionisation energyA. `Ne lt F`B. `O gt N`C. `Na gt AI`D. `Mg gt AI` |
| Answer» Correct Answer - D | |
| 219. |
The amount of energy required to remove the most loosely bound electron from an isolated gaseous atom is called as first ionization energy `(IE_(1))`. Similarly the amount of energies required to knock out second, third etc. electrons from the isolated and `IE_(3)gt IE_(2)gt IE_(1)`. (i) Nuclear charge (ii) Atomic size (iii) penetration effect of the electrons (iv) shielding effect of the inner electrons and (b) electronic configurations (exactly half filled and completely filled configurations are extra stable) are the important factors which affect the ionisation energies. Similarly, the amount of energy released when a neutral isolated gaseous atom accepts an extra electron to from gaseous anion is called electron affinity. `(X(g)+e^(-)(g)rarr X^(-)(g)+` energy A positive elecrton affinity idicates that the ion `X^(-)` has a lower more negative energy than the neutral atom X. The second electron affinity for the addition of a second electron to an initially neutral atom is negative because the electron replusion outweights the nuclear attraction, e.g., `O(g)+e^(-)overset("Exothermic")rarr O^(-)(g),E_(a)=+141 kJ mol^(-)` ....(i) `O^(-)(g)+e^(-)overset("Excothermic")rarr, E_(a)=-780 kJ mol^(-)` ...(ii) The electron affinity of an element depends upon (i) atomic size (ii) nuclear charge and (iii) electronic configuration. In general, in a group, ionisation energy and electron affinity decrease as the atomic size increases. The members of third period have some higher (e.g., S and Cl) electron affinity values than the members of second period (e.g., O and F) because second period elements have very small atomic size. Hence, there is tendency of electron-electron repulsion, which resultss in less evolution of energy in the formation of correcsponding anion. Which one of the following statements is incorrect in relation to ionisation enthalpy?A. Ionization enthalpy increase for each successive electron.B. The greatest increase in ionization enthalpy is experienced on removal of electron from core of noble gas configurationC. End of valence electrons is marked by a big jump in ionization enthalpyD. Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value. |
| Answer» Correct Answer - D | |
| 220. |
The amount of energy required to remove the most loosely bound electron from an isolated gaseous atom is called as first ionization energy `(IE_(1))`. Similarly the amount of energies required to knock out second, third etc. electrons from the isolated and `IE_(3)gt IE_(2)gt IE_(1)`. (i) Nuclear charge (ii) Atomic size (iii) penetration effect of the electrons (iv) shielding effect of the inner electrons and (b) electronic configurations (exactly half filled and completely filled configurations are extra stable) are the important factors which affect the ionisation energies. Similarly, the amount of energy released when a neutral isolated gaseous atom accepts an extra electron to from gaseous anion is called electron affinity. `(X(g)+e^(-)(g)rarr X^(-)(g)+` energy A positive elecrton affinity idicates that the ion `X^(-)` has a lower more negative energy than the neutral atom X. The second electron affinity for the addition of a second electron to an initially neutral atom is negative because the electron replusion outweights the nuclear attraction, e.g., `O(g)+e^(-)overset("Exothermic")rarr O^(-)(g),E_(a)=+141 kJ mol^(-)` ....(i) `O^(-)(g)+e^(-)overset("Excothermic")rarr, E_(a)=-780 kJ mol^(-)` ...(ii) The electron affinity of an element depends upon (i) atomic size (ii) nuclear charge and (iii) electronic configuration. In general, in a group, ionisation energy and electron affinity decrease as the atomic size increases. The members of third period have some higher (e.g., S and Cl) electron affinity values than the members of second period (e.g., O and F) because second period elements have very small atomic size. Hence, there is tendency of electron-electron repulsion, which resultss in less evolution of energy in the formation of correcsponding anion. The first ionisation energy of `Na, Mg,AI` and `Si` are in the order of:A. `Na lt Mg gt AI lt Si`B. `Na gt Mg gt AI gt Si`C. `Na lt Mg lt AI lt Si`D. `Na gt Mg gt AI gt Si` |
| Answer» Correct Answer - A | |
| 221. |
The outermost electronic configuration of most electropositive element isA. `ns^(1)`B. `ns^(2)np^(2)`C. `ns^(2)np^(3)`D. `ns^(2)np^(5)` |
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Answer» Correct Answer - A |
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| 222. |
The ionisation energy of nitrogen is more than that of oxygen becauseA. of the extra stability of half-filled p orbitals in nitrogenB. of the smaller size of nitrogenC. The former contains less number of electronsD. The former is less electronegative |
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Answer» Correct Answer - A |
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| 223. |
The `I_(1)` values of `Li,Be` and C are `5.4 eV//"atom", 9.32 eV//"atom"` and `11.26 eV//"atom"`. The `I_(1)` value of Boron isA. `13.6eV//"atom"`B. `8.29 eV//"atom"`C. `14.5eV//"atom"`D. `21.5eV//"atom"` |
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Answer» Correct Answer - B |
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| 224. |
The ionisation enthalpy of lithium is much lower than that of helium despite the fact that the nuclear charge of lithium is `+3` and that of helium is `+2`. It is due to:A. 2s electron in lithium is more strongly bound to the nucleus than the 1s electron in HeB. 2s electron in lithium is equally bound to the nucleus than the 1s electron in HeC. 2s electron in lithium is less strongly bound to the nucleus than the 1s electron in HeD. Atomic size of He is more than Li |
| Answer» Correct Answer - C | |
| 225. |
The less electropositive element isA. NaB. BeC. LiD. Mg |
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Answer» Correct Answer - B |
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| 226. |
Electropositivity is very high forA. AlB. GeC. LiD. Ba |
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Answer» Correct Answer - D |
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| 227. |
The first ionisation energies of Li, Be, B and C are in the order:A. (a) `Li gt Be lt B lt C`B. (b) `Li lt Be gt B lt C`C. (c) `Li gt Be gt B gt C`D. (d) `Li lt Be gt B gt C` |
| Answer» Correct Answer - B | |
| 228. |
The ionisation energy of Li is 500 KJ/mole. The amount of energy required to convert 70 mg of Li atoms required to convert 70 mg of Li atoms in gaseous state into `Li^(+)` ion (in KJ) is |
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Answer» Correct Answer - 5 |
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| 229. |
The correct order of the second ionisation potential of carbon, nitrogen, oxygen and fluorine isA. `C lt N lt O lt F`B. `C lt O lt N lt F`C. `C lt N lt F lt O`D. `C lt O lt F lt N` |
| Answer» Correct Answer - C | |
| 230. |
Second ionisation potential of `Li,Be,B` is in the order:A. `Li gt Be gt B`B. `Li gt B gt Be`C. `Be gt Li gt B`D. `B gt Be gt C` |
| Answer» Correct Answer - B | |
| 231. |
The second ionisation enthalpies of elements are always higher than their first ionisation enthalpies because:A. the cation is smaller than its parent atomB. it is easier to remove electron from cationC. ionization is an ednothermic processD. cation formed always have stable half filled or completely filled valence shell electron configuration |
| Answer» Correct Answer - A | |
| 232. |
The second ionisation energy of N and O in electron volt are respectively given by:A. 29, 29B. 34, 34C. 29, 34D. 34, 29 |
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Answer» Correct Answer - C |
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| 233. |
The first ionisation potential in electron volts of nitrogen and oxygen atoms are respectively given byA. `14.6, 13.6`B. `13.6,14.6`C. `13.6,13.6`D. `14.6,14.6` |
| Answer» Correct Answer - A | |
| 234. |
The first ionisation potential of `Na,Mg,Al` and `Si` are in the orderA. `Na lt Mg gt AI lt Si`B. `Na gt Mg gt AI gt Si`C. `Na lt Mg lt AI lt Si`D. `Na gt Mg gt AI lt Si` |
| Answer» Correct Answer - A | |
| 235. |
The first ionisation potential of AI is smaller than that of Mg because:A. the atomic size of `AI gt Mg`B. the atomic size of `Ai lt Mg`C. AI has one unpaired electron in 3p-orbitalD. Mg has incompletely filled 3s-orbital |
| Answer» Correct Answer - C | |
| 236. |
The charge/size ratio of a cation determines its polarising power. Which one of the following sequeces represents the increasing order of the polarising power of the cationic species, `K^(+),Ca^(2+),Mg^(2+),Be^(2+)`?A. `Mg^(2+) lt Be^(2+) lt K^(+) lt Ca^(2+)`B. `Be^(2+) lt K^(+) lt Ca^(2+) lt Mg^(2+)`C. `k^(+) lt Ca^(2+) lt Mg^(2+) lt Be^(2+)`D. `Ca^(2+) lt Mg^(2+) lt Be^(2+) lt K^(+)` |
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Answer» Correct Answer - C |
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| 237. |
In which of the following compound size of cation to anion ratio is minimum?A. `CsF`B. `LiI`C. `LiF`D. `CsI` |
| Answer» Correct Answer - B | |
| 238. |
The number of different possible oxidation states of fluorine in the molecules [`F_(2),OF_(2),O_(2)F_(2),HF] = t` Find the value of t |
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Answer» Correct Answer - 2 `0, -1` |
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| 239. |
Assertion :- Cr belong to 4th period Reason :- Cr atom contain 4 shells.A. If both Assertion & Reason are True & the Reason is a correct explanation of the Assertion.B. If both Assertion & Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True, but the Reason is False.D. If both Assertion & Reason are false |
| Answer» Correct Answer - A | |
| 240. |
In the sixth period, the orbitals being filled areA. `5s,5p,5d`B. `6s,6p,6d,6f`C. `6s,5f,6d,6p`D. `6s,4f,5d,6p` |
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Answer» Correct Answer - D |
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| 241. |
The elements in which electrons are progressively filled in 4f-orbitals are calleD:A. actinoidsB. transition elementsC. lanthanidesD. halogens |
| Answer» Correct Answer - C | |
| 242. |
The correct decreasing order of size of a isoelectronic species is `-`A. `Se^(-2) gt Br^(-) gt Kr gt Rb^(+) gt Sr^(+2)`B. `S^(-2) gt Cl^(-) gt K^(+) gt Ar gt Ca^(+2)`C. `N^(-3) gt O^(-2) gt Ne gt F^(-)Ca^(+2)`D. `F^(-) gt Ne gt Na^(+) gt Al^(+3) gt Mg^(+2)` |
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Answer» Correct Answer - 1 Size of isoelectronic species `prop (e)/(P)"ratio , "e/P" "overset(Se^(-2))(36/34) gt overset(Br^(-))(36/35) gt overset(Kr)(36/36) gt overset(Rb^(+))(36/37) gt overset(Sr^(+2))(36/38)` |
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| 243. |
The ionic radius of `Cr` is minimum in which of the following compounds ?A. `CrF_(3)`B. `CrCI_(3)`C. `Cr_(2)O_(3)`D. `K_(2)CrO_(4)` |
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Answer» Correct Answer - D `Cr` has maximum oxidation number `(+6)` in `K_(2)CrO_(4)` and thus, has minimum ionic radius. |
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| 244. |
Write the correct formula and name of the following (a) `ICI` or `CII` (b) `FCI` or `CIF` (c) `BrCi` or `CiBr` (d) `Brl` or `IBr` (e) `OF_(2)` or `F_(2)O` (f) `CI_(2)O` or `OCI_(2)` |
| Answer» `{:("Correct formula","Name"),((a)I^(+)CI^(-),"Iodine chloride"),((b)CI^(+)F^(-),"Chlorine fluoride"),((c)Br^(+)CI^(-),"Bromine chloride"),((d)IBr,"Iodine bromide"),((e)OF_(2),"Oxygen difluoride"),((f)CI_(2)O,"Dichlorine oxide"):}` | |
| 245. |
Which one of the following statements is incorrect in relation to ionisation enthalpy?A. Ionization enthalpy increases for each successive electron.B. The greatest increase in ionization enthalpy is experienced on removal of electron from core of noble gas configuration.C. End of valence electrons is marked by a big jump in ionization enthalpy.D. Removal of electron from orbitals bearing lower `n` value is easier than from orbitals having higher `n` value. |
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Answer» Correct Answer - D Orbitals bearing lower value of n will be more closer to the nucleus and thus electrons will experience greater attraction from nucleus and so its removal will be difficult, not easier. |
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| 246. |
Which orbitals are filled in given order in 5th period of periodic table?A. `5s rarr 4d rarr 5p`B. `5s rarr 4d rarr 5d`C. `4s rarr 5d rarr 4p`D. `5s rarr 3d rarr 4p` |
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Answer» Correct Answer - A According to `n+l` rule ectrons will be filled in the following order of orbitals, `5s rarr 4d rarr 5p`. |
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| 247. |
`K^(+),CI^(-),Ca^(2+),S^(2-)` ions are isoelectronic. The decreasing order of other size is:A. `S^(2-), Br^(-), K^(+), Ca^(2+)`B. `Br^(-),S^(2-),K^(+),Ca^(2+)`C. `K^(+),Ca^(2+),S^(2-),Br^(-)`D. `Ca^(2+),K^(+),S^(2-),Br^(-)` |
| Answer» Correct Answer - B | |
| 248. |
`K^(+),CI^(-),Ca^(2+),S^(2-)` ions are isoelectronic. The decreasing order of other size is:A. `S^(2-) gt CI^(-) gt K^(+) gt Ca^(2+)`B. `Ca^(2+) gt K^(+) gt CI^(-) gtS^(2-)`C. `K^(+) gt CI^(-) gt Ca^(2+) gt S^(2-)`D. `CI^(-) gt S^(2-) gt Ca^(2+) gt K^(+)` |
| Answer» Correct Answer - A | |
| 249. |
The first ionisation energy of Al is smaller than that of Mg because `:`A. The atomic number of Al is smaller than that of Mg.B. the atomic size of Al is less than that of Mg.C. Penetration of `s-` subshell electrons in case of Mg is greater than that of `p-` subshell in AlD. Mg has incompletely filled `s-` orbital. |
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Answer» Correct Answer - 3 Penetration of `p-` subshell electron is less than `s-` subshell electrons. In case of `Mg`, the first electron is to be removed from completely filled `3s^(2)` valence shell configuration as compared to partially filled `3p^(1)` of Al. These two factors collectively accounts for the higher ionisation energy of Mg than that of Al. Therefore, `(C )` option is correct. |
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| 250. |
The first ionisation potential of `Na` is `5.1eV`. The value of eectrons gain enthalpy of `Na^(+)` will beA. `-2.55 eV`B. `-5.1 eV`C. `-10.2 eV`D. `+2.55 eV` |
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Answer» Correct Answer - B `{:(Na rarr Na^(+) +e^(-),,I^(st)I.E. = 5.1 eV),(Na^(+) rarr+e^(-) rarr Na,,"Electron gain enthalpy of" Na^(+)):}` Because reaction is reverse, so `Delta_(eg)H =- 5.1 eV`. |
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