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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. |
An element with atomic number 20 will be placed in which period of the periodic tableA. 5thB. 4thC. 3rdD. 2nd. |
| Answer» Correct Answer - B | |
| 302. |
There are four elements P,Q,R and S having atomic numbers of 4,18,10 and 16 respectively. The element which can exhibit covalency as well as electrovalency will be :A. PB. QC. RD. S |
| Answer» Correct Answer - D | |
| 303. |
The atomic numbers of four elements A,B,C and D are 12,13, 15 and 3 respectively. The element which cannot form a cation is :A. AB. BC. CD. D |
| Answer» Correct Answer - C | |
| 304. |
Ionisation energy of `He^+` is `19.6 xx 10^-18 J "atom"^(-1)`. The energy of the first stationary state `(n = 1)` of `Li^( 2 +)` is. |
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Answer» Correct Answer - `-4.41xx10^(-17) J "atom"^(-1)` lonisation energy of `He^(+)` ion =`19.6xx10^(-18) J "atom"^(-1)` Energy of first orbit (ground state) of `He^(+)` ion =E_(prop)-l.E.`=0-19.6xx10^(-18)=-19.6xx10^(-18) J "atom"^(-1)` Energy of first stationary state (ground state) o f`Li^(+)` ion =`(-19.6xx10^(-18)xx(Z_(Li))^(2))/((Z_(He))^(2))` `=-19.6xx10^(-18)xx(9)/(4)=-44.1xx10^(-18)J "atom"^(-1)=-4.41xx10^(-17) J "atom"^(-1)`. |
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| 305. |
Ionisation energy of `He^+` is `19.6 xx 10^-18 J "atom"^(-1)`. The energy of the first stationary state `(n = 1)` of `Li^( 2 +)` is.A. `8.82xx010^(-17) J "atom"^(-1)`B. `4.41xx10^(-16) J "atom"^(-1)`C. `-4.41xx10^(-17) J "atom"^(-1)`D. `-2.2xx10^(-15) J "atom"^(-1)` |
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Answer» Correct Answer - C `IE=-E_(1)` `E_(1) "for" He^(+) "si" =-19.6xx10^(-18) "J mol"^(-1)` `((E_(1))_(He^(+)))/((E_(1))_(Li^(2+)))=((Z_(He^(+)))^(3))/((Z_(Li^(2+)))^(2))` `(-19.6xx10^(-18))/((E_(1))_(Li^(2+)))=(4)/(9)` `E_(1_(Li^(2+)))=((-19.6xx10^(-18))xx9)/(4)` `=-4.41 "J atom"^(-1)` |
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| 306. |
If an electron travels with a velocity of 1/100th speed of light in the first in the first Bohr orbit, what is its velocity (relative to the speed of light) in the 5th Bohr orbit ?A. 0.002B. 0.1C. 0.5D. 0.7 |
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Answer» Correct Answer - A `v_(n) = v_(0) (Z)/(n)` `v_(1) = (1)/(100) xx c = v_(0) (1)/(1) = v_(0), " i.e., " v_(0) = (c)/(100)` `v_(5) = v_(0) (1)/(5) = (c)/(100) xx (1)/(5) = (c)/(500) = 0.002 c` |
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| 307. |
Ionisation energy of `He^(+) " is " 19.6 xx 10^(-18)J "atom"^(-1)`. The energy of the first stationary state `(n = 1)` of `Li^(2+)` isA. `-2.2 xx 10^(-15) J "atom"^(-1)`B. `8.82 xx 10^(-17) J "atom"^(-1)`C. `4.41 xx 10^(-16)J "atom"^(-1)`D. `-4.41 xx 10^(-17) J "atom"^(-1)` |
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Answer» Correct Answer - D `E_(n) = - (2pi^(2) m Z^(2) e^(4))/(n^(2) h^(2)) = - K (Z^(2))/(n^(2))` I.E. of `H^(+) = E_(oo) - E_(1) = 0 - (-K (2^(2))/(1^(2)))` `= 4K = 19.6 xx 10^(-18) J "atom"^(-1)` (Given) or `K = 4.9 xx 10^(-18) J "atom"^(-1)` For `Li^(2+) , Z = 3` and for first stationary state, n = 1 `:. E_(1) = - K (Z^(2))/(n^(2)) = - 4.9 xx 10^(-18) xx (3^(2))/(1^(2))` `= -4.41 xx 10^(-17) J "atom"^(-1)` |
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| 308. |
For the electrons of oxygen atom, which of the following statemetns correct?A. `Z_(eff)` for an electron in a 2s orbital is the same as `Z_(eff)` for an electron in a 2p orbtialB. An electron in the 2s orbtial has the same energy as an electron in the 2p orbitalC. `z_(eff)` for an electron in 1s orbital is the same as `z_(eff)` for an electron in a 2s orbitalD. The two electrons present in the 2s orbtial have spin quantum numbers `m_(s)` but of opposite sign. |
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Answer» (a) Electrons in 2s and 2p orbtials have different screen effect. Hence their `Z_(eff)` is different. `Z_(eff)` of 2s orbital `gtZ_(eff)` of 2p orbital (b) Energy of 2s orbital `lt` energy of 2p orbtial Hence, it is not correct. (c) `Z_(eff)` of 1sorbital`cancel=Z_(eff)` 2s orbital Hence, it is incorrect. (d) For the two electrons of 2s orbtial, the value of `m_(s)` is `+(1)/(2)` and `-(1)/(2)` hence, it is correct. |
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| 309. |
The reaction between iron and copper sulphate solution represents which type of reaction ?A. decompositionB. combinationC. single displacementD. double decomposition |
| Answer» Correct Answer - C | |
| 310. |
When a student added a few drops of barium chloride solution to sodium sulphate solution, he obtained a white precipitate instantly. Which of the following type of chemical reaction has been carried out by the student ?A. combinationB. double displacementC. displacementD. decomposition |
| Answer» Correct Answer - B | |
| 311. |
We can show that is more reactive than copper :A. by preparing copper sulphate solution and dipping iron strip in itB. by dipping both the strips in water for some timeC. by preparing iron sulphate solution and dipping copper strip in itD. by heating both iron and copper strips |
| Answer» Correct Answer - A | |
| 312. |
Match the following : `{:((i) "X-rays",(a) v = 10^(0) - 10^(4) Hz,),((ii) UV,(b) v = 10^(10) Hz,),((iii) "Long radio waves",(c) v = 10^(16) Hz,),((iv) "Microwaves",(d) v = 10^(18) Hz,):}` |
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Answer» Correct Answer - (i)-(d), (ii)-(c), (iii)-(a), (iv)-(b) v values are in the order : X-rays `gt UV gt` Microwaves `gt` Radiowaves |
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| 313. |
The atomic number of an element is derived from theA. number of electronB. number of protonsC. number of neutronsD. number of isotopes |
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Answer» Correct Answer - B Ato0mic number of an element is defined as the number of protons in the nucleus of the atom of the element. |
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| 314. |
Alpha `(alpha)` particles areA. `He` atomsB. `He^(2+)`ionsC. `He^(+)`ionsD. `He^(-)`ions |
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Answer» Correct Answer - B Alpha particles are `He^(2+)` ions, i.e. `He`atoms without their two electrons. It was confirmed by the fact that alpha particles yielded `He` gas when they combined with two electrons. |
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| 315. |
The orbital angular momentum of a p-electron is given as:A. `sqrt((3)/(2)) (h)/(pi)`B. `sqrt6 (h)/(2pi)`C. `(h)/(sqrt2pi)`D. `sqrt3 (h)/(2pi)` |
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Answer» Correct Answer - C For p-obrbital, `l = 1` Orbital angular momentum `= sqrt(l(l + 1)) (h)/(2pi) = sqrt(1(1 + 1)) (h)/(2pi)` `= sqrt2 (h)/(2pi) =(h)/(sqrt2pi)` |
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| 316. |
For `d` electron, the orbital angular momentum isA. `sqrt(6)((h)/(2pi))`B. `sqrt(2)((h)/(2pi))`C. `((h)/(2pi))`D. `2((h)/(2pi))` |
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Answer» Correct Answer - A According to quantum mechanics, the orbital angular momentum is given as `sqrt(l(l+1)).(h)/(2pi)` For a `d` electron, `l = 2`, thus, `mvr = sqrt(2(2+1)) (h)/(2pi) = sqrt(6) ((h)/(2pi))` |
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| 317. |
Studies on the phenomenon of radioactivity supported the conclusion that the (i) atom was divisible (ii) atom could spilt into netural particles (iii) atom could split into charged particles (iv) atom was indivisibleA. `(i),(iii)`B. `(ii),(iii)`C. `(i)`D. `(iv)` |
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Answer» Correct Answer - A The emission of `alpha` and `beta` particles implied the subdivision of atom into charged partiles. |
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| 318. |
We all know that a large number of elements exist as isotopes. The isotopes of an element have same atomic number but different mass numbers. The isotopes. With higher mass number are mostly radioactive in nature. These are called radioisotopes. These isotopes have a wide range of applications. (i) Which isotope is used to study the kinetics of photosynthesis ? (ii) Name the isotope which can detect a blood clot. (iii) Which isotope is used to detect the deformity in bones, if any ? |
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Answer» (i) `.^(14)C` isotope is used in this case. (ii) `.^(24)Na` isotope can detect the blood clot. (iii) `.^(32)P` isotope can detect the deformity in bones. |
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| 319. |
(i)The uncertainties in the potisition and velocity of particle are `3.14 xx 10^(-10)m` and`5.27 xx 10^(-24) m//s` respectively . Calculate mass of the particle. (ii)Find the numberof waves made by a Bohr electron in one complete revolution in the 3rd Bohr orbit. |
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Answer» (i) As `Delta x . Deltap = (h)/(4pi)` `impliesDelta x . mDelta v = (h)/( 4pi)` `:. m = ( h)/( 4pi xx Deltax xx Deltav) = ( 6.6 xx 10^(-34))/( 4 xx 3.14 xx 10^(-10) xx 5.27 xx 10^(-24)) = 10kg` (ii) Circumference of 3rd orbit `=2pi r_(3)` `mvr_(3) = ( 3h)/( 2pi) = ( 3h)/( 2pi) ` ....(A) For de-Brogile equation `lambda= ( h)/( mv)` ....(B) From equation (A) & (B) `lambda= ( h)/( 3h // 2pi r_(3)) implies3 lambda = 2pir_(3)` or `2pir_(3) 3 lambda` . Hence, 3 waves are formed in one complete revolution. |
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| 320. |
Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other types of material. If the velocity of the electron in this microscope is `1.6 xx 10^(6) ms^(-1)`, calculate de Broglie wavelength associated with this electron. |
| Answer» `lamda = (h)/(mv) = (6.626 xx 10^(-34) kg m^(2) s^(-1))/((9.11 xx 10^(-31) kg) (1.6 xx 10^(6) ms^(-1))) m = 4.55 xx 10^(-10) m = 455` pm | |
| 321. |
g sub-shell is characterised by :A. l=5B. l=3C. l=4D. l=5 |
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Answer» Correct Answer - C For g-subshell, l=4. |
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| 322. |
Which electronic configuration is not allowed for a neutral atom or an ion in ground state ?A. `1s^(2) 2s^(2) 2p^(6) 3s^(1)`B. `1s^(2) 2s^(2) 2p^(6) 3d^(5)`C. `1s^(2) 2s^(2) 2p^(6) `D. `1s^(2) 2s^(2) 2p^(6) 3s^(2)` |
| Answer» Correct Answer - 2 | |
| 323. |
The wavelength of `H_(alpha)` line of Balmer series is `6500 Å`. What is the wavelength of `H_(beta)` line of Balmer series? |
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Answer» `H_(alpha)` line of Balmer series is obtained when `n_(1) = 2, n_(2) = 3` `H_(beta)` line of Balmer series is obtained when `n_(1) = 2, n_(2) = 4` Thus, `bar(v) H_(alpha) = (1)/(.^(lamda)H_(alpha)) = R_(H) ((1)/(2^(2)) - (1)/(3^(2))) = R_(H) ((1)/(4) - (1)/(9)) = R_(H) xx (5)/(36)` ...(i) `bar(v) H_(beta) = (1)/(.^(lamda)H_(beta)) = R_(H) ((1)/(2^(2)) - (1)/(4^(2))) = R_(H) ((1)/(4) - (1)/(16)) = R_(H) xx (3)/(16)`..(ii) Dividing eqn. (i) by eqn. (iii), we get `(.^(lamda)H_(beta))/(.^(lamda)H_(alpha)) = (5)/(36) xx (16)/(3) = (20)/(27) :. .^(lamda)H_(beta) = (20)/(27) xx .^(lamda)H_(beta) = (20)/(27) xx 6500 Å = 4814.8 Å` |
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| 324. |
1 mole of photons, each of frequency `250 s^(-1)` would have approximately a total energy ofA. 1 ergB. 1 jouleC. 1 eVD. 1 MeV |
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Answer» Correct Answer - A `E = N_(A) hv = (6.02 xx 10^(23)) (6.6 xx 10^(-34) Js) (250 s^(-1)) = 10^(-7) "erg " = 1J` |
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| 325. |
The number of d-electrons in `Fe^(2+)`(Z=26) is not equal to the number of electrons in which one of the following ?A. d-electrons in Fe (Z=26)B. p-electrons in Ne (Z=10)C. s-electrons in Mg(Z=12)D. p-electrons in Cl(Z=17) |
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Answer» Correct Answer - D The electronic configurations of the atoms involved are : (a) Ff(Z=26) , `[Ar]^(18)3d^(6) 4s^(2) " " ` (d electrons = 6) (b) Ne(Z=10) , `1s^(2)2s^(2)2p^(6) " "` (p electrons = 6) (c) (Mg=12) , `1s^(2)2s^(2)2p^(6)3s^(2) " "` (s electrons =6) (d) (Cl=17) , `1s^(2) 2s^(2)2p^(6) 3s^(2)3p^(5) " "` (p electrons =11) `Fe^(2+)` ion has six d electrons and these are not equal to p-electron (11) in chlorine (Cl) atom. |
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| 326. |
A particle of charge `q` and mas `m` si accelerated from set through a potential difference `V`. Its de Broglie wavelength is equal toA. `sqrt((h)/(2mq V))`B. `(hqV)/(sqrt(2m))`C. `sqrt((hqV)/(2m))`D. `(h)/(sqrt(2mqV))` |
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Answer» Correct Answer - D When a charge `q` is accelerated from rest through a potential difference `V`, its gain in kinetic enegry `(1)/(2)mV^(2)` must equal its loss in potential enegry `qV`, since the energy is conserved. That is, `(1)/(2)mV^(2) = qV` Since momentum, `p = mV`, we can express this in the form `(1)/(2)(m^(2)v^(2))/(m) = qV rarr (p^(2))/(2m) = qV` or `p = sqrt(2mqV)` Substituting this expression for `p` into the de Broglie relation `(lambda = h//p)` gives `lambda = (h)/(sqrt(2mqV))` |
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| 327. |
In photoelectric effect, the kinetic energy of the photoelectrons increase linearly with theA. wavelength of incident lightB. frequency of incident lightC. velocity of incident lightD. atomic mass of the element |
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Answer» Correct Answer - B `KE = hv - hv_(0)` |
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| 328. |
Calculate the ratio between the wavelength of an electron and a proton if the proton is moving with half the velocity of electron (mass of proton `= 1.67 xx 10^(-27)kg` and mass of electron `= 9.11 xx 10^(-31) kg`) |
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Answer» Correct Answer - 916.6 `lamda_(e) = (h)/(m_(e) v_(e)), lamda_(p) = (h)/(m_(p) v_(p)), (lamda_(e))/(lamda_(p)) = (m_(p) v_(p))/(m_(e) v_(e)) = (1)/(2) xx (1.67 xx 10^(-27) kg)/(9.11 xx 10^(-31) kg) = 916.6` |
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| 329. |
Which of the following correctly represent the electronic distribution in the Mg atom?A. 3,8,1B. 2,8,2C. 1,8,3D. 8,2,2 |
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Answer» Correct Answer - b correct answer b |
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| 330. |
If an atom contains one electron and one proton, will it carry any charge or not? |
| Answer» No, the atom will not carry any electric charge. This is because the electron has 1 unit negative charge whereas a proton has an equal and opposite 1 unit positive charge due to which the net charge on the atom is zero. | |
| 331. |
What do you think would be the observation if the `alpha`-particle scattering experiment is carried out using a foil of a metal other than gold ? |
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Answer» (i) If the foil of a heavy metal like platinum (having a very heavy nucleus) is used, then the observations in the `alpha`-particle scattering experiment would be the same as that in the gold foil experiment (ii) If, however, the foil of a right metal like lithium (having a very light nucleus) is used, then the fast moving heavy `alpha`-particles may even push the light nucleus aside and may not be deflected back. |
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| 332. |
With the help of Table given below, find out the mass numbers of oxygen and sulphur atoms : |
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Answer» (i) Mass number of oxygen = No. of protons + No. of neutrons `= 8 +8` `=16` (ii) Mass number of sulphur = No. of protons + No. of neutrons `= 16 + 16` `= 32`. |
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| 333. |
Which of the following statements about Rutherford model of atom are correct?A. iand iiiB. ii and iiiC. I and ivD. only I |
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Answer» Correct Answer - a Correct answer (a) The statement ii and iii are both correct |
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| 334. |
There are two species represented as `.^(35)Cl and .^(37)Cl`. Which of the following statement is correct regarding these species ?A. they have different chemical propertiesB. their physical properties are the sameC. they have the same number of protonsD. they are isobars of the same element |
| Answer» Correct Answer - C | |
| 335. |
The number of electrons in an element X is 15 and the number of neutrons is 16. Which of the following is the correct representation of the element ?A. `overset(31)underset(15)X`B. `overset(31)underset(16)X`C. `overset(16)underset(15)X`D. `overset(15)underset(16)X` |
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Answer» Correct Answer - a correct answer a Mass no = o of electrons (protons)+no of neutrons =15+16=31 |
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| 336. |
The number of electrons in the atom of an element X is 15 and the number of neutrons is 16. Which of the following is the correct representation of an atom of this element ?A. `._(15)^(31)X`B. `._(16)^(31)X`C. `._(15)^(16)X`D. `._(16)^(15)X` |
| Answer» Correct Answer - A | |
| 337. |
Which of the following elements does not exhibit electrovalency ?A. calciumB. chromiumC. carbonD. cadmium |
| Answer» Correct Answer - C | |
| 338. |
Which of the following statement is correct about the atom of an element ?A. an atom can have only protons and neutrons but no electronsB. an atom can have only electrons and neutrons but no protonsC. an atom can have only electron and proton but no neutronD. an atom must always have a proton, neutron and electron |
| Answer» Correct Answer - C | |
| 339. |
The mass number of two atoms X and Y is the same (40 each) but their atomic numbers are different (being 20 and 18 respectively). X and Y are examples of :A. chemically similar atomsB. isotopesC. solid and liquid metalsD. isobars |
| Answer» Correct Answer - D | |
| 340. |
Which of the following statement is always correct?A. an atom has equal number of electrons and protonsB. an atom has equal number of electrons and neutronsC. an atom has equal number of protons and neutronsD. an atom has equal number of electrons, protons and neutrons. |
| Answer» Correct Answer - A | |
| 341. |
For`n=4` , which one of the following values of l is not possible ?A. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - 4 | |
| 342. |
Which one of the following orbitals is spherical in shape ?A. 4sB. 3p-oribitalC. 3dD. 4f |
| Answer» Correct Answer - 1 | |
| 343. |
The first model of an atom was given byA. Neils BohrB. Ernest RutherfordC. J.J. ThomsonD. Eugen Goldstein |
| Answer» Correct Answer - C | |
| 344. |
Four elements W,X,Y and Z contain 8,11,9 and 17 protons per atom respectively. The element which cannot form an anion is most likely to be :A. WB. XC. YD. Z |
| Answer» Correct Answer - B | |
| 345. |
The number of valence electrons in a graphite atom is :A. 2B. 4C. 3D. 5 |
| Answer» Correct Answer - B | |
| 346. |
The number of valence electrons in Aluminium isA. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - 3 | |
| 347. |
The number of valence electrons in a sulphide ion, `S^(2-)` is :A. 16B. 10C. 9D. 8 |
| Answer» Correct Answer - D | |
| 348. |
The four atomic species can be represented as follows. Out of these, the two species which can be termed isobars are : (i) `._(60)^(201)X` (ii) `._(61)^(200)X` (iii) `._(58)^(200)X` (iv) `._(60)^(203)X`.A. (i) and (ii)B. (ii) and (iii)C. (i) and (iii)D. (i) and (iv) |
| Answer» Correct Answer - B | |
| 349. |
The first model of an atom was given byA. N. BohrB. E.GoldsteinC. RutherfrodD. J.J Thomson |
| Answer» Correct Answer - d | |
| 350. |
Number of valence electrons in `Cl^(– )` ion are: |
| Answer» A chlorine atom (Cl) has 7 valence electrons. A chloride ion is formed by the addition of 1 more electron to a chlorine atom. So, the number of valence electron in a chloride ion `(Cl^(-))` will be `7 + 1=8`. | |