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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.
| 301. |
Which of the following have linear shapes?A. `SaCI_(2)`B. `NO_(2)^(+)`C. `FNO`D. `SO_(2)` |
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Answer» Correct Answer - b `NO_(2)^(+)` is lone `O = Nrarr O` |
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| 302. |
A bond is said to have ionic character when the electronegativitly differnce between the two atoms is greater than ………. |
| Answer» Correct Answer - 1.9 | |
| 303. |
Number of `sigma` bonds , `pi` bonds and lone pair on Xe form of `XeOF_(4)`A. `4,1,1`B. `4 , 2, 1`C. `5,1,1`D. `6, 2, 0` |
| Answer» Correct Answer - C | |
| 304. |
Bond order of `O_(2), O_(2)^(-) and O_(2)^(2-)` is in orderA. `O_(2) gt O_(2)^(+) gt O_(2)^(2-) gt O_(2)^(-)`B. `O_(2)^(-) gt O_(2)^(2-) gt O_(2)^(+) gt O_(2)`C. `O_(2)^(+) gt O_(2) gt O_(2)^(-) gt O_(2)^(2-)`D. `O_(2)^(2-) gt O_(2)^(-) gt O_(2) gt O_(2)^(+)` |
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Answer» Correct Answer - c Given species `O_(2) O_(2)^(-),O_(2)^(+)` and `O_(2)^(2-)` Total number of electron `{:(O_(2)rarr16e^(-)O_(2)^(-1),,17e^(-),O_(2)^(+1)rarr15e^(-),O_(2)^(-2)rarr,18e^(-),,),(,,O_(2)^(+1),O_(2),O_(2)^(-1),O_(2)^(-2),,),("Bond order",,2.5,2,1.5,1,,):}` Higher bond order indicates greater stability Stability order `[O_(2)^(+1) gt O_(2) O_(2)^(-1) gt O_(2)^(2+)]` |
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| 305. |
Water has high heat of vaporisation due to ?A. covelent bondingB. H- bondingC. ionic bondingD. none of the above |
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Answer» Correct Answer - b Hydrogen bonding increasing heat of vaprisation . |
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| 306. |
Percentage of p- character in each orbital of centain atom used bonding in `NH_(3)` isA. `25%`B. `75%`C. More than `75%`D. `33.3%` |
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Answer» Correct Answer - c When B.A is `109^(@)28^(@) = 75%` p-character in orbital If `B.A. is 107^(@)=` more than ` 75%` p-character in orbital . |
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| 307. |
The correct order of bond order values among the following : `A . NO^(-) B.NO^(+) C.NO D.NO^(2+) E. NO^(2-)` isA. `A lt D lt C lt B lt E`B. `D = B lt A lt E le C `C. `E lt A lt D = C lt B`D. `B lt C lt D lt A lt E ` |
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Answer» Correct Answer - C `NO=KK sigma_(2)^(2) sigma_(2s)^(**2) sigma_(2p_(z))^(2) pi_(2p_(x))^(2)pi_(2p_(y))^(2)pi_(2p_(x))^(**1)` B.O. `= (1)/(2) (8-3) = (5)/(2) = 2.5 ` `NO^(-)=KK sigma_(2)^(2) sigma_(2s)^(**2) sigma_(2p_(z))^(2) pi_(2p_(x))^(2)pi_(2p_(y))^(2)pi_(2p_(x))^(**1)pi_(2p_(y))^(**1)` B.O. = `(1)/(2) (8-4)= 2` `NO^(2-)=KK sigma_(2)^(2) sigma_(2s)^(**2) sigma_(2p_(z))^(2) pi_(2p_(x))^(2)pi_(2p_(y))^(2)pi_(2p_(x))^(**2)pi_(2p_(y))^(**1)` B.O. = `(1)/(2) (8-5) = (3)/(2) = 1.5 ` `NO^(+)=KK sigma_(2)^(2) sigma_(2s)^(**2) sigma_(2p_(z))^(2) pi_(2p_(x))^(2)pi_(2p_(x))^(2)pi_(2p_(y))^(2)` B.O. `= (1)/(2) (8-2) = 3 ` `NO^(2+)=KK sigma_(2)^(2) sigma_(2s)^(**2) sigma_(2p_(z))^(2) pi_(2p_(x))^(2)pi_(2p_(y))^(1)` B.O. ` = (1)/(2) (7-2) = (5)/(2) = 2.5 ` Thus `NO^(2-) lt NO^(-) NO = NO^(2-) lt NO^(+)`, i.e. `E lt A lt C = D lt B ` |
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| 308. |
Which of the following statements is not correct form the view point of molecular orbital theory ?A. `Be_(2)` is not a stable moleculeB. `He_(2)` is not stable but `He_(2) ^(+)` is expected to existC. Bond strength of `N_(2)` is maximum amongst to homonuclear diatomic molecular belonging to the second periodD. The order of energies of molecular obitals in `N_(2)` molecule is ` sigma 2slt sigma^(**) 2s lt sigma 2p_(z) lt (pi 2p_(x) = pi 2p_(y))` `lt (pi^(**) 2p_(x) = pi^(**) 2p_(y)) lt sigma^(**) 2p_(z)` |
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Answer» Correct Answer - D The correct order of energies of molecular orbitals in `N_(2)` molecule is `sigma _(2s) lt sigma_(2s)^(**) lt (pi_(2_(p_(x))) = pi_(2_(p_(y)))) lt sigma _(2_(p_(z)))lt (pi_(2_(p_(x)))^(**) = pi_(2_(p_(y)))^(**)) lt sigma_(2_(p_(z)))^(**)` |
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| 309. |
Which of the following molecular species has unpaired electrons(s) ? .A. `N_(2)`B. `F_(2)`C. `O_(2)^(-)`D. `O_(2)^(2)` |
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Answer» Correct Answer - c `sigma 1s^(2), sigma ** 1s^(2), sigma 2s^(2), sigma** 2s^(2), sigma p_(x)^(2) pi 2p_(y)^(2), pi 2p_(z)^(2), pi ** 2p_(y)^(2) pi ** 2p_(z)^(1)` Hence , `O_(2)^(-)` has unpaired electron. |
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| 310. |
Among the following species, which has the minimum bond length?A. `B_(2)`B. `C_(2)`C. `F_(2)`D. `O_(2)^(-)` |
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Answer» Correct Answer - b `B_(2)` bond order `= 1,C_(2)` bond `= 2, F_(2)` bond order `= 1`, `O_(2)^(-)` bond order for 1.5, bond order` = 1//`bond length |
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| 311. |
which one of the following pairs consists of only paramagnetic species ?A. `[O)_(2), NO]`B. `[O_(2)^(+) , O_(2)^(2-)]`C. `[CO, NO]`D. `[NO, No^(+)]` |
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Answer» Correct Answer - A Both `O_(2)` and NO are paramagnetic as `O_(2)` contains two unpaired electrons and NO has one unpaired electron. |
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| 312. |
Which of the following species is paramagnetic ? .A. `CO_(2)`B. `O_(2)^(2-)`C. `CN^(Theta)`D. `NO` |
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Answer» Correct Answer - D In `NO` there is one unpaired electron in antibonding `pi MO` . |
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| 313. |
Paramagnetic species areA. `O_(2)^(+)`B. `O_(2)^(-)`C. `N_(2)^(+)`D. `N_(2)^(-)` |
| Answer» Correct Answer - A::B::C::D | |
| 314. |
Which of the following has the minimum bond length ?A. `O_(2)`B. `O_(2)^(+)`C. `O_(2)^(-)`D. `O_(2)^(2-)` |
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Answer» Correct Answer - B `O_(2)^(+)` has highest B. O . |
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| 315. |
Among the following , the paramagnetic compound is :A. `Na_(2)O_(2)`B. `O_(3)`C. `N_(2)O`D. `KO_(2)` |
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Answer» Correct Answer - d `Na_(2)O_(2) is (2Na) (O_(2)^(2))`. Thus is no unpaired either on `Na^(+) or O_(2)^(2)`. Hence it must be diamagnetic `O_(3) and N_(2)O` have even numbered electrons and thus are diamagnetic |
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| 316. |
Ionic compounds in general possess bothA. high melting points and non-directional bondsB. high melting points and low boiling pointsC. directional bonds and low boiling pointsD. high solubilities in polar and non-polar bonds |
| Answer» Correct Answer - A | |
| 317. |
Which atomic orbital is always involved in sigma bonding only ?A. pB. dC. sD. none |
| Answer» Correct Answer - C | |
| 318. |
The electronic configuration of four elements `L, P, Q` and `R` are given in brackets `L (1s^(2), 2s^(2), 2p^(4)) , P (1s^(2), 2p^(6), 3s^(1))` `Q (1s^(2), 2s^(2)2p^(6), 3s^(2)3p^(5)) , R (1s^(2), 2s^(2)2p^(6), 3s^(2))` The formula of ionic compounds that can be formed between elements areA. `L_(2)P, RL, PQ and R_(2)Q`B. `LP, RL, PQ and RQ`C. `P_(2)L, RL, PQ and RQ_(2)`D. `LP, R_(2)L, P_(2)Q and RQ` |
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Answer» Correct Answer - c Valencies of `L, Q, P and is -2, -1, +1 and +2` respectively so they will form `P_(2)L, RL, PQ and RQ_(2)`. |
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| 319. |
The electronic configuration of four elements `L, P, Q` and `R` are given in brackets `L (1s^(2), 2s^(2), 2p^(4)) , P (1s^(2), 2p^(6), 3s^(1))` `Q (1s^(2), 2s^(2)2p^(6), 3s^(2)3p^(5)) , R (1s^(2), 2s^(2)2p^(6), 3s^(2))` The formula of ionic compounds that can be formed between elements areA. `L_2P, RL, PQ, R_2Q`B. LP, RL, PQ, RQC. `P_2L, RL, PQ, RQ_2`D. `LP, R_2L,P_2Q, RQ` |
| Answer» Correct Answer - C | |
| 320. |
The electronic configuration of metal M is `1s^(2)2s^(2)2p^(6)3s^(1).` The formula of its oxide will be :A. MOB. `M_(2)o`C. `M_(2)O_(3)`D. `MO_(3)` |
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Answer» Correct Answer - B Metal is monovalent and its oxide is `M_(2)O.` |
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| 321. |
You are given the electronic configuration of five neutral atoms - AB, B, C, D and E ` A - 1s^(2) 2s^(2) 2p^(6) 3s^(2), B - 1s^(2), 2s^(2) 2p^(6) 3s^(1), C - 1s^(2) 2s^(2) 2p^(1) , D- 1s^(2) 2s^(2) 2p^(5), E - 1s^(2) 2s^(2) 2p^(6)` Write the empirical formula for the substance containing (i) A and D (ii) B and D (iii) only D (iv) only E ? |
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Answer» (i) Empirical formula of the compound formed by A and D `AD_(2)` as A has two valence electrons and D has seven. Atom A transfers its two electrons to two D atoms to complete their octets . (ii) Empirical formula of the compound formed by B and D is BD as B transfers its one electron to D. (iii) `D_(2)` as both the atoms of D share one electron each to form a covelent bond (iv) Since it is noble gas , no compound is formed . |
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| 322. |
The values of electronegativity of atoms A and B are 1.2 and 4.0 respectivity . The % ionic character of the A-B bond isA. `50%`B. `72.24%`C. `55.3%`D. `43%` |
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Answer» Correct Answer - B According to Hannay and Smith equation, % ionic character = `16 (chi_(A) - chi_(B)) + 3.5(chi_(A) - chi_(B))^(2)` ` 16 (4-0 - 1.2) + 3.5 (4 .0 - 1.2)^(2)` `= 16 xx2.8 + 3.5 xx (2.8)^(2)` `= 44.8 + 27 .44 = 72.24` |
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| 323. |
The correct statement with regard to `H_(2)^(+) and H_(2)^(-)` isA. both`H_(2) ^(+) and H_(2)^(-)` do not existB. `H_(2)^(-) ` is more stable than ` H_(2)^(+)`C. `H_(2)^(+)` is more stable than ` H_(2)^(-)`D. both ` H_(2)^(+) and H_(2)^(-)` are equally stable |
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Answer» Correct Answer - C B. O . Is same but ` H_(2) ^(-)` has one eletron is ABMO. |
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| 324. |
For the formation of an ionic bond between two atoms, one atom should have …………….. And the other atom should have ………… |
| Answer» ionization enthalpy , high eletron gain enthalpy | |
| 325. |
Arrange in order of increasing strength of hydrogen bonding : `O, F, S, Cl, N` |
| Answer» `S lt Cl lt N lt O lt F`. | |
| 326. |
The number of electrons shared by each N atom in `N_2` isA. 2B. 1C. 3D. 4 |
| Answer» Correct Answer - C | |
| 327. |
Why `H_(2) O ` is a liquid while `H_(2) S` is a gas ? |
| Answer» In `H_(2) O`, there is hydrogen bonding and hence association of `H_(2) O ` molecular but in `H_(2) S` there is no H- bonding . | |
| 328. |
Why ethyl alcohol is competely miscible with water ? |
| Answer» This is because ethyl alcohol forms H-bonds with water. | |
| 329. |
Why HF has higher boiling point than HCl ? |
| Answer» In HF , there is H-bonding and hence the molecules are associated together but in HCl, there is no H-bonding . | |
| 330. |
`H_(2)O` has higher boiling point than `H_(2)S` becauseA. `H_(2)S` is a smaller molecule and hence more closely packedB. the bond angle of `H_(2)O` is more than `H_(2)S` and hence `H_(2)O` molecule are more tightly packetC. the intermolecular hydrogen bonding in liquid `H_(2)O`D. the latent head of voparisation is higher for `H_(2)O` than for `H_(2)S` |
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Answer» Correct Answer - c The intermolecular bonding in liquid `H_(2)O`. |
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| 331. |
The total number of `pi` bond electrons in the following structure is A. 12B. 16C. 4D. 8 |
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Answer» Correct Answer - d There are double bonds hence no of pui- electrons `= 2 xx 4 = 8` |
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| 332. |
The species, having bonds angle of `120^(@)` isA. `CIF_(3)`B. `NCl_(3)`C. `BCl_(3)`D. `PH_(3)` |
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Answer» Correct Answer - c a,b,d have irregular geometry due to lone pair so bond angle not fixed `BCI_(3)` has `sp^(2)` hybridisation bond angle `= 120^(@)` |
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| 333. |
Comprehension given below is followed by some multiple choice question, Each question has one correct options. Choose the correct option. Molecular orbitals are formed by the overlap of atomic orbitals. Two atomic orbitals combine to form two molecular orbitals called bonding molecular orbital (BMO) and anti-bonding molecular orbital (ABMO). Energy of anti-bonding orbital is raised above the parent atomic orbitals that have combined and hte energy of the bonding orbital is lowered than the parent atomic orbitals. energies of various molecular orbitals for elements hydrogen to nitrogen increase in the order `sigma1s lt sigma^(star)1s lt sigma^(star)2s lt ((pi2p_(x))=(pi2p_(y))) lt sigma2p_(z) lt (pi^(star)2p_(x) = pi^(star)2p_(y)) lt sigma^(star)2p_(z)` and For oxygen and fluorine order of enregy of molecules orbitals is given below. `sigma1s lt sigma^(star)1s lt sigma2s lt sigma^(star)2s lt sigmap_(z) lt (pi2p_(x) ~~ pi2p_(y)) lt (pi^(star)2p_(x)~~ pi^(star)2py) lt sigma^(star)2p_(z)` Different atomic orbitalsof one atom combine with those atoms orbitals of the second atom which have comparable energies and proper orientation. Further, if the overlapping is head on, the molecular orbital is called sigma, `sigma` andif the overlap is lateral, the molecular orbital is called pi, `pi`. The molecular orbitals are filled with electrons according to the same rules as followed for filling of atomic orbitals. However, the order for filling is not the same for all molecules or their ions. Bond order is one of the most important parameters to compare the strength of bonds. 67) Which of the following pair is expected to have the same bonod order?A. `O_(2),N_(2)`B. `O_(2)^(+), N_(2)^(-)`C. `O_(2)^(-), N_(2)^(+)`D. `O_(2)^(-), N_(2)^(-)` |
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Answer» Correct Answer - b b) On the basec of molecular orbital theory we can calculated bond order of molecules ions as `BO=1/2(N_(b)-N_(a))` Molecular orbital electronic configuration (MOEC) of `N_(2)` is `sigma1s^(2), sigma^(star)1s^(2), sigma2s^(2), sigma^(star)2s^(2), pi2p_(x)^(2)~~pi2p_(y)^(2), sigma2p_(x)^(2)` Bond order of `N_(2) = 1/2(10-4)=3` MOEC of `N_(2)^(+) = sigma1s^(2),sigma^(star)1s^(2), sigma2s^(2), sigma^(star)2s^(2), pi2p_(x)^(2) ~~ pi2p_(Y)^(2), sigma2p_(2)` BO of `N_(2)^(+) =1/2(9-4)=2/5` MOEC of `N_(2)^(-)=sigma1s^(2),sigma^(star)1s^(2),sigma2s^(2), sigma^(star)2s^(2), pi2p_(x)^(2)~~ pi2p_(y)^(2), sigma2p_(z)^(2), pi^(star)2p_(x)^(1)~~ pi^(star)2p_(y)` BO of `N_(2)^(-)=1/2(10-5)=2.5` MOEC of `O_(2)=sigma1s^(2), sigma^(star)1s^(2), sigma2s^(2), sigma^(star)2s^(2), sigma2p_(z)^(2), pi2p_(x)^(2)~~ pi2p_(y)^(2), pi^(star)2p_(x)^(1)~~ pi^(star)2p_(y)^(1)` BO of `O_(2)=1/2(10-6)=2` MOEC of `O_(2)^(-)=sigma1s^(2), sigma^(star)1s^(2),sigma2s^(2),sigma^(star)2s^(2),sigma2p_(z)^(2), pi2p_(x)^(2)~~pi2p_(y)^(2), pi^(star)2p_(x)^(2)~~pi^(star)2p_(y)^(1)` BO of `O_(2)^(-) = 1/2(10-7)=15` MOEC of `O_(2)^(+) = sigma1s^(2), sigma^(star)1s^(2),sigma2s^(2),sigma^(star)2s^(2),sigma2p_(z)^(2),pi2p_(x)^(2)~~pi2p_(y)^(2), pi^(star)2p_(x)^(2)~~pi^(star)2p_(y)` BO of `O_(2)^(+)=1/2(10-5)=2.5` a) Bond order of `O_(2)` and `N_(2)` are 2 and 3 respectively. b) Bond order of both `O_(2)^(+)` and `N_(2)^(-)` are 2.5 c) Bond order of `O_(2)^(-)` and `N_(2)^(-)` are 1.5 and 2.5 respectively. |
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| 334. |
The energy of hydrogen bond is of the order ofA. `2 kJ mol^(-1)`B. `20 kJ mol^(-1)`C. `200 kJ mol^(-1)`D. `2000 kJ mol^(-1)` |
| Answer» Correct Answer - B | |
| 335. |
The molecule which has highest bond order isA. `C_2`B. `N_2`C. `B_2`D. `O_2` |
| Answer» Correct Answer - B | |
| 336. |
Which of the following oxides is not expected to react with sodium hydroxide ?A. `BeO`B. `B_(2)O_(3)`C. `CaO`D. `SiO_(2)` |
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Answer» Correct Answer - c CaO is basic oxide. |
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| 337. |
What is meant by the term bond order ? Calculate the bond order ? Calculate the bond order of : `N_(2),O_(2),O_(2)^(-)and O_(2)^(2-)` |
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Answer» For the difinition of bond order m cinsult section 4.11 `{:(overset(* *)N-=overset(* *)N,overset(* *)":O"=overset(* *)O":",overset(-)C-=overset(+)O":"),("B.O.=3","B.O.=2",".B.O.=3"):}` |
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| 338. |
What is meant by the term bond order? Calculate the bond order of `N_(2)`,`O_(2)`,`O_(2)^(o+)` and `O_(2)^(ө)`. |
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Answer» Bond order is defined as one half of the difference between the number of electrons present in the bonding and anti-bonding orbitals of a molecule. If `N_(a)` is equal to the number of electrons in an anti-bonding orbital, then `N_(b)` is equal to the number of electrons in a bonding orbital. Bond order `=(1)/(2)(N_(b)-N_(a))` If `N_(b) gt N_(a)`, then the molecule is said be stable. However , if `N_(b) le N_(a)`, then the molecule is considered to be unstable. Bond order of `N_(2)` can be calculated form its electronic configuration as : `[sigma(1s)]^(2)[sigma^(**)(1s)]^(2)[sigma(2s)]^(2)[sigma^(**)(2s)]^(2)[pi(2p_(x))]^(2)[pi(2p_(y))]^(2)[sigma(2p_(z))]^(2)` Number of bonding electron , `N_(b)=10` Number of anti-bonding electron , `N_(a)=4` Bond order of nitrogen molecule =`(1)/(2)(10-4)=3` There are 16 electrons in a dioxygen molecule, 8 from each oxygen atom. The electronic configuration of oxygen molecule can be written as : `[sigma -(1s)]^(2)[sigma^(**)(1s)]^(2)[sigma(2s)]^(2)[sigma^(**)(2s)]^(2)[sigma(1p_(z))]^(2)[pi(2p_(x))]^(2)[pi(2p_(y))]^(2)[pi^(**)(2p_(x))]^(1)[pi^(**)(2p_(y))]^(1)` Since the 1s orbital of each oxygen atom is not involved in bonding, the numebr of bonding electrons=8=`N_(b)` and the number of anti-bonding electrons =4=`N_(a)` Bond order `=(1)/(2)(N_(b)-N_(a))` `=(1)/(2)(8-4)=2` Hence, the bond order of oxygen molecule is 2. Similarly, the electronic configuration of `O_(2)^(+)` can be written as : `KK[sigma(2s)]^(2)[sigma^(**)(2s)]^(2)[sigma(2p_(z))]^(2)[pi(2p_(x))]^(2)[pi(2p_(y))]^(2)[pi^(**)(2p_(x))]^(1)` `N_(b)=8` `N_(a)=3` Bond order of `O_(2)^(+)=(1)/(2)(8-3)=2.5` Thus, the bond order of `O_(2)^(+)` is 2.5 The electronic configuration of `O_(2)^(-)` ion will be : `KK[sigma(2s)]^(2)[sigma^(**)(2s)]^(2)[sigma(2p_(z))]^(2)[pi(2p_(x))]^(2)[pi(2p_(y))]^(2)[pi^(**)(2p_(x))]^(2)[pi^(**)(2p_(y))]^(1)` `N_(b)=8` `N_(a)=5` Bond order of `O_(2)^(-)=(1)/(2)(8-5)=1.5` Thus, the bond order of `O_(2)^(-)` ion is 1.5. |
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| 339. |
In the following question a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below each eqestion. Assertion (A) : Sodium chloride formed by the action of chlorine gas on sodium metal is a stable compound Reason (R) : This is because sodium and chloride ions acquire octet in sodium chloride formation.A. A and R both are correct, and R is the correct ecplanation of A.B. A and R both are correct, But R is not the correct ecplanation of A .C. A is true but R is false .D. A and R both are false. |
| Answer» Correct Answer - A | |
| 340. |
Some of the properies of the two species `NO_(2)^(-)` and `H_(3)O` are described below which one of them is correct ?A. Dissimilar in hybridization for the central atom with different atomB. Isostructual with the same hybridization for the central atomC. Isostructural with the difference hybridization for the central atomD. Similar in hydridization for the central atom with defferent structure. |
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Answer» Correct Answer - a Appliying the formula `X = (1)/(2) [V.E + MA - C + a]` For `NO_(3)^(-),X = (1)/(2) [5 + 0 - 1] = 3` Hybridisation `= sp^(2)` , structure planar triangular For `H_(3)O^(+) , X = (1)/(2) [6 + 3 - 1 + 0]= 4` `rArr sp^(3)` hybridisation , pyramidal structure .Thus hybridisation and the structure are different. |
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| 341. |
The stability of ions of Ge Sn and `Pb` will be in the order .A. `Ge^(2+) lt Sn^(2+) lt Pb^(2+)`B. `Pb^(2+) gt Pb^(4+)`C. `Sn^(4+) gt Pb^(2+)`D. `Ge^(4+) lt Sn^(4+) lt Pb^(4+)` |
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Answer» Correct Answer - A::B Down the group stability of lower oxidation state increases due to inert pair effect `underset("Decreasing stability")overset(Pb^(2+) gt SN^(2+) gt Ge^(2+))rarr` Also `pb^(2+)` is more stable than `Pb^(4+)` . |
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| 342. |
Assertion . The bond anlge of `PBr_(3)` is greater than that of `PH_(3)` but bond angle of `NBr_(3)` is less than that of `NH_(3)`. Reason .Electronegativity of phosphorus atom is less than that of nitrogen .A. If both assertion and reason are correct, and reason is the correct explanation of the assertion.B. If both assertion and reason are correct , but reason is not the correct explanation of the assertion.C. If assertion is correct, but reason is incorrect .D. If both assertion and reason are incorrect. |
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Answer» Correct Answer - b Correct explanation. The truth of assertion is not due to difference in the electronegativities of N and P but due to difference in electronegativity of H and Br (in `NH_(3) and NBr_(3))` and due to reaonance in ` PBr_(3)` but no resonance in `PH_(3)` |
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| 343. |
Which of the following is the most ionic ? .A. `P_(4)O_(10)`B. `MnO`C. `CrO_(3)`D. `Mn_(2)O_(7)` |
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Answer» Correct Answer - B In `overset(+2)(Mn) overset(-2)O_(2)` the `OS` of Mn is +2 and is the lowest Hence most ionic `overset(+5)(P_(4)) O_(10) overset(+6)CrO_(3), overset(+7)(Mn)_(2) O_(7)` . |
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| 344. |
Name one compound each involving `sp^(3) , sp^(2) and sp ` hybridisation. |
| Answer» `sp^(3) = CH_(4), sp^(2) = C_(2)H_(4), sp = C_(2) H_(2).` | |
| 345. |
How does velence bond theory, explain the existence of `H_(2)` but non-existence of `H_(2)`? |
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Answer» According to valence bond theory, a covalent bond is formed by overlap of half-filled atomic orbitals : H-atoms have half-filled orbitals `(1s^(1))` but He atoms have fully filled orbitals `(1s^(2))`. |
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| 346. |
Predicted the correct order among the followingA. Ione pair -line pair gt bond pair - bond gt lone pair - lone pairB. Ione pair -lone pair gt lone pair - bond pair gt bond pair - bond pairC. Ione pair -lone pair gt bond pair - bond pair gt lone pair - bone pairD. bond pair -bond pair gt lone pair - bond pair gt Ione pair - Ione pair |
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Answer» Correct Answer - b The order of repulsion force according to VSEPR theory lone pair -lone gt lone pair -bond pair gt bond paiur - bond pair |
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| 347. |
Which out of `N_(2)` and `H_(2)O` is polar and why ? |
| Answer» Nitrogen `(N_(2))` is a homodiatomic molecule and is therefore non-polar. On the other hand, the geometry of `H_(2)O` is angular or bent with bond angle of `104.5^(@).` The polarities of the O-H bonds do not cancel out and it is of polar nature. | |
| 348. |
Which properties of element depend on the electronic configuration of atoms and which do not (b) Why the chemical properties of a group are similar Why do they not have identical properties (c ) Account for the great chemical similarity of the lanthanoid elements (Z =57 to 71) (d) Select the largest species in each group (i) `Ti^(2),Ti^(+3)` (ii) `F^(Θ),Ne,Na^(o+)` (e) Select the species with the largest `IE` in each group (i )`na,K,Rb` (ii) F,Ne,Na (f) Which ion has the smallest radius `Li^(o+),Na^(o+),K^(o+) Be^(+2), Mg^(+2)` (g) Select among the element that has the lowest and highest IE `K, Ca, Se, B, Kr`, (h) The ionic radii of `S^(2-)` and `Te^(2-)` are and 220 pm respectively Predict the ionic radius of `Se^(2-)` and for `p^(3-)` (i) In the ionic compound `KF` the `K^(o+)` and `F^(o+)` have practically identical radii, about 134 pm each. Predict the relative atomic radii of 1`K` and `F` . |
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Answer» Cemical and many physical properties depend on electonic cofiguration, but nuclear properties do not (b) Their properties are similar because of similar outermost electrnic configurations . The differnces are due to their differences in size and the magnitude of shielding effect (c) Their outermost two shells of electrons are alike `(6s^(2)5d^(1-))` or that configuration is very close to the ground state in stability Moreover all the the lanthanides act as if they had this configuration to a great extent (d) Ti (ii) `F` (e) Na (ii) Ne (f) `Be^(2-)` Be is a second period element, thus smaller than the higher period elements. The `Be^(2+)` ion is isoelectronic with `Li^(o+)` but has a greater nuclear charge (g) K has the lowest `IE` Kr has the highest IE increases along period `(rarr)` being very small for alkali metals and very large for the noble gases (h) Ionic radius of `Se^(2-)` is the average of those of `S^(2-)` and `Te^(2-)` that is `202 `pm The actual value is `198 pm` The expected radius of `P^(3-)` to be greater than that of `S^(2-)` since two ions are isoelectronic but `P` has one less proton Its actual radius is 212 pm (i) The atomic radius of `K` should be much greater than 134Pm and that of `F` much smaller since atomic cations are smaller than their parent atoms, while atomic anions are bigger than their parents The observed atomic radius of `K` and `F` are `231` and 64 pm respectively . |
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| 349. |
A bond formed between two similar atoms cannot beA. ionicB. co-ordinateC. covalentD. `pi`-bond |
| Answer» Correct Answer - A | |
| 350. |
During change of `NO^(+)to NO`, the electron is added toA. `sigma`-orbitalB. `pi`-orbitalC. `sigma^**` -orbitalD. `pi^**`-orbital. |
| Answer» Correct Answer - D | |