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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.
| 801. |
If the differentiating electron enters (n-1) d-sublevel. The element isA. A representiative elemetB. A noble gasC. An alkali metalD. A transition element |
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Answer» Correct Answer - D |
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| 802. |
In iron atom `(z=26)`, the differentiating electron enters ….. SublevelA. 4dB. 3dC. 4pD. 5p |
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Answer» Correct Answer - B |
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| 803. |
Lanthanides are group of elements in which the differentiating electron enters intoA. s-sub levelB. d-sub levelC. p-sub levelD. f-sub level |
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Answer» Correct Answer - D |
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| 804. |
The correct sequence which shows decreasing order of the ionic radii of the elements isA. `AI^(3+) gt Mg^(2+) gt Na^(+) gt F^(-) gt O^(2-)`B. `Na^(+) gt Mg^(2+) gt AI^(3-) gt O^(2-) gt F`C. `Na^(+) gt F^(-) gt Mg^(2+) gt O^(2-) gt AI^(3+)`D. `O^(2-) gt F^(-) gt Na^(-) gt Mg^(2+) gt AI^(3-)` |
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Answer» Correct Answer - D For isoelectronic species, ionic radii `prop (1)/("nuclear charge")`. So, correct order of ionic radii is `._(8)O^(2-) gt _(9)F^(-) gt _(11)Na^(+) gt _(12)Mg^(2+) gt _(13)AI^(3+)` |
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| 805. |
The magnitude of electron affinity depends uponA. Electron affinityB. Polarising powerC. Ionization potentialD. The nuclear charge |
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Answer» Correct Answer - D |
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| 806. |
The outer electronic configuration of Gd (At.No. 64) isA. `4f^(3)5d^(5)6s^(2)`B. `4f^(8)5d^(6)6s^(2)`C. `4f^(4)5d^(4)6s^(2)`D. `4f^(7)5d^(1)6s^(2)` |
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Answer» Correct Answer - D Gadolinium `(._(64)Gd)) = [Xe]^(54) 4f^(7) 5d^(1)6s^(2)` |
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| 807. |
Ionisation energy depends upon:A. Principle qunatum numberB. Azimuthal quantum numberC. Magnetic quantum numberD. (1) & (2) both |
| Answer» Correct Answer - D | |
| 808. |
Atomic radius depends uponA. anionic natureB. nature of bondingC. cation natureD. metalic nature |
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Answer» Correct Answer - B |
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| 809. |
Lanthanoid contraction is caused due to:A. The imperfect shielding on outer electrons by 4f-electrons from the nuclear chargeB. The appreciable shielding on outer electrons by 4f-electrons from the nuclear chargeC. The apperciable shielding on outer electrons by 5d-electrons from nuclear chargeD. The same effective nuclear charge from `Ce` to `Lu` |
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Answer» Correct Answer - A |
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| 810. |
The percentage ionic character in H-Br bond in HBr molecule is `"________________"`, given the electronegativity of H and Br are 2.1 and 2.8 respectively .A)`12.9`B)`16.9`C)`9`A. `12.9`B. `16.9`C. `23.9`D. `9` |
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Answer» Correct Answer - A |
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| 811. |
The percentage ionic character in H-Br bond in HBr molecule is `"________________"`, given the electronegativity of H and Br are 2.1 and 2.8 respectively .A. `12.9`B. `16.9`C. `23.9`D. `9` |
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Answer» Correct Answer - A |
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| 812. |
Among elements with the following electromic configurations, the one with the largest radius isA. `[Ne] 3s^(2)`B. `[Ne] 3s^(2) 3p^(1)`C. `[Ne] 3s^(2) 3p^(3)`D. `[Ne] 3s^(2) 3p^(5)` |
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Answer» Correct Answer - A |
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| 813. |
The first ionization potentials of four consecutive elements present in the second period of periodic table are 8.3, 11.3, 14.5, and 13.6 eV respectively which one of the following is the first ionization potential of nitrogen ?A. 13.6B. 11.3C. 8.3D. 14.5 |
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Answer» Correct Answer - D |
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| 814. |
For univalent elements, the average value of first ionization potential and first electron affinity is equal to itsA. Polarising powerB. Covalent radiusC. ElectronegativityD. Dipole moment |
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Answer» Correct Answer - C |
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| 815. |
`A_(2)` and `B_(2)` are two diatomic molecules with bond energies of A-A and B-B bonds as x and y respectively. If the bond energy of the molecule, A-B formed up from `A_(2)` and `B_(2)` is z. then, the responance energy of molecules A-B will beA. `(DeltaE)_(A-B) = Z - sqrt (xy)`B. `(DeltaE)_(A-b) = x-y-z`C. `(DeltaE)_(A-b) = z-x+y`D. `(DeltaE)_(A-b) = sqrt (xy-z)` |
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Answer» Correct Answer - A |
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| 816. |
Which of the following is correct statement?A. The members of second period have lower electron affinity than the next member in their respective groups of p-blockB. The electron affinity of Si is greater than that of PC. `F^(-)` is strongest reducing agentD. Tl, Pb, & Bi exhibit inert pair effect. |
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Answer» Correct Answer - C |
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| 817. |
The electronic configuration of four elements are given below. Arrange these elements in the correct order of the magnitude of their electron affinity. `(i)2s^(2)p^(3)" "(ii)3s^(2)3p^(5)" "(iii)2s^(2)2p^(1)" "(iv)3s^(2)3p^(4)` Select the correct answer using the codes given below `:`A. `(ii) lt (i) lt (iv) lt (iii)`B. `(i) lt (iii) lt (iv) lt (ii)`C. `(iii) lt (iv) lt (ii) lt (i)`D. `(iii) lt (iv) lt (i) lt (ii)` |
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Answer» Correct Answer - 2 Order of electron affinity `rarrClgtSgtOgtN` |
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| 818. |
An element of atomic mass `40` has `2,8,8,2`, as the electronic configuration. One of the following statement regarding this elements is not correct?A. it forms an basic oxideB. It belongs to II A groupC. It belongs to IV periodD. It forms an acidic oxide |
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Answer» Correct Answer - 4 The element is calcium and hence its oxide `(CaO)` is basic in nature. |
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| 819. |
Element with electronic configuration as `[Ar]^(18)d^(5)4s^(2)` is placed in `:`A. `1^(st)` group , `s-` blockB. `2^(nd)` group, `s-` blockC. `5^(th)` group, `d-` blockD. `7^(th)` group, `d-` block |
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Answer» Correct Answer - 4 Last electrons enters in `d-` orbital, So it belong to `d-` block. For `d-` block elements the group number is equal to the number of valence electrons `+` number of electrons in `(n-1)d` subshell `i.e.,2+5=7^(th)` group. |
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| 820. |
Element in periodix table with electronic configuration as `[Ar]^(18)3d^(5)4s^(1)` is placed in:A. IA,s-blockB. VIA,s-blockC. VIB,s-blockD. VIB,d-block |
| Answer» Correct Answer - D | |
| 821. |
The size of the species, `Pb,Pb^(2+), Pb^(4+)` decreases as -A. `Pb^(4+) gt Pb^(2+) gt Pb`B. `Pb gt Pb^(2+) gt Pb^(4+)`C. `Pb gt Pb^(4+) gt Pb^(2+)`D. `Pb^(4+) gt Pb gt Pb^(2+)` |
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Answer» Correct Answer - B `[(z)/(e) uarr` Atomic size `darr`] |
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| 822. |
Bond distance `C-F` in `(CF_(4))` & `Si-F` in `(SiF_(4))` are respective `1.33 Å` & `1.54 Å. C-SI` bond is `1.87 Å`. Calculate the colvalent radius of F atom ignoring the electronegativity differences.A. `0.64 Å`B. `(1.33 +1.54 +1.8)/(3) Å`C. `0.5 Å`D. `(1.54)/(2) Å` |
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Answer» Correct Answer - C `[r_(c) +r_(F) = 1.33 Å` `underset(-----------)(r_(Si)+r_(F) = 1.54 Å)` `r_(c) +r_(Si) +2r_(F) = 2.87 Å` `r_(c) +r_(Si) = 1.87 Å` `1.87 Å + 2r_(F) = 2.87 Å` `2r_(F) = 1.00 Å` `r_(F) = 5.Å]` |
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| 823. |
Consider the following sequence of reaction : If electronic configuration of element X is `[Ne]3s^(1)` , then which of the following order is correct regarding given enthalpies ?A)`|DeltaH_(4)| = |DeltaH_(5)|`B)`|DeltaH_(2)| gt |DeltaH_(1)|`C)`|DeltaH_(2)| gt |DeltaH_(3)|`D)`|DeltaH_(1)| = |DeltaH_(6)|`A. `|DeltaH_(4)| = |DeltaH_(5)|`B. `|DeltaH_(2)| gt |DeltaH_(1)|`C. `|DeltaH_(2)| gt |DeltaH_(3)|`D. `|DeltaH_(1)| = |DeltaH_(6)|` |
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Answer» Correct Answer - A::B::D |
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| 824. |
Which of the following are correct .A. `IP_(1)` of He `lt IP_(2)` of LiB. The element with highest `IP_(2)` is LithiumC. All d-block elements are transition elementsD. Across a transition series from left to right the IP values increase slowly . |
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Answer» Correct Answer - A::B::D |
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| 825. |
Consider the following sequence of reaction : If electronic configuration of element X is `[Ne]3s^(1)` , then which of the following order is correct regarding given enthalpies ?A. `|DeltaH_(4)| = |DeltaH_(5)|`B. `|DeltaH_(2)| gt |DeltaH_(1)|`C. `|DeltaH_(2)| gt |DeltaH_(3)|`D. `|DeltaH_(1)| = |DeltaH_(6)|` |
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Answer» Correct Answer - A::B::D |
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| 826. |
Highest oxidation states shown by Chromium & Maganese are `+x & +y` respectively. Give the value of `x +y`? |
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Answer» Correct Answer - 13 `+6, +7` `6 +7 = 13` |
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| 827. |
Atomic radius of `Li is 1.23 Å` and ionic radius of `Li^(+)` is `0.76 Å`. Calculate the percentage of volume occupied by single valence electron in `Li`. |
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Answer» Volume of `Li = (4)/(3) xx 3.14 xx (1.23)^(3) = 7.79 Å (-Li = 1s^(2) 2s^(1))` Volume of `Li^(+) = (4)/(3) xx 3.14 xx (0.76)^(3) = 1.84 Å (Li^(+) = 1s^(2))` `:.` Volume occupied by `2s` subsheel `= 7.79 - 1.84 = 5.95 Å`. `%` volume occupied by single valence electron i.e., `2s` electron `= (5.95)/(7.79) xx 100 =76.4%`. |
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| 828. |
If internuclear distance between `A` atoms in `A_(2)` is `10 Å` and between `B` atoms in `B_(2)` is `6Å`, then calculate internuclear distance between `A` and `B` in `Å` [Electronegativity difference between `A` and `B` has negligible value]. |
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Answer» Correct Answer - 8 `r_(A)+r_(A) =10 Å ..(i)` `r_(B) +r_(B) = 6Å …(ii)` (i) +(ii) `2(r_(A)+r_(B)) = 16Å` `r_(A) +r_(B) = 8Å` |
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| 829. |
In which of the following arrangements , the order is not correct according to the property indicated against it:A. Increasing size `Al^(3+) lt Mg^(2+) lt Na^(+) lt F^(-)`B. Increasing `IE_(1): B lt C lt N lt O`C. Increasing `EA_(1):I lt Br lt F lt Cl`sD. Increasing metallic radius: `Li lt Na lt K lt Rb` |
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Answer» Correct Answer - B |
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| 830. |
Which of the following processes involve absorption of energy?A. `S(g)+e^(-) rarr S^(-)(g)`B. `Xe(g)+e^(-) rarr Xe^(-)(g)`C. `O^(-)(g)+e^(-) rarr O^(2-)(g)`D. `CI^(-)(g) rarr CI(g)+e^(-)` |
| Answer» Correct Answer - B::C::D | |
| 831. |
which of the following order of ionisation energy is correct?A. `Li gt Be (II I.E.)`B. `N gt O (II I.E.)`C. `Be lt B (II I.E.)`D. `F lt O (II I.E.)` |
| Answer» Correct Answer - A::C::D | |
| 832. |
In which of the following arrangements, the order is not correct according to the property indicated against it.a)increase size :`Cu^(2+) lt Cu^(+) lt Cu`b)increasing `IE_(1): B lt C lt N lt O`c)increasing `IE_(1) : Na lt Al lt Mg lt Si`d)increasing `IE_(1) :Li lt Na lt K lt Rb`A. increase size :`Cu^(2+) lt Cu^(+) lt Cu`B. increasing `IE_(1): B lt C lt N lt O`C. increasing `IE_(1) : Na lt Al lt Mg lt Si`D. increasing `IE_(1) :Li lt Na lt K lt Rb` |
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Answer» Correct Answer - B::D |
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| 833. |
In which of the following arrangements, the order is not correct according to the property indicated against it.A. Increasing size: `Cu^(2+) lt Cu^(+) lt Cu`B. Increasing `IE_(1): B lt C lt N lt O`C. Increasing `IE_(1): Na lt AI lt Mg lt Si`D. Increasing `IE_(1): Li lt Na lt K lt Rb` |
| Answer» Correct Answer - B::D | |
| 834. |
In which of the following arrangements the order is NOT according to the property indicated against it ?A. `Ai^(3+) lt Mg^(2+) lt Na^(+) lt F^(-)` increasing ionic sizeB. `B lt C lt N lt O lt P`-increasing first ionisation enthalpyC. `I lt Bt lt F lt CI`increasing electron gain enthalpy (with negative sign)D. `Li lt Na lt K lt Rb`-increasing metallic radius |
| Answer» Correct Answer - B | |
| 835. |
In which of the following arrangements the order which is correct according to the property indicated against it.A. `AI^(+3) lt Mg^(+2) lt Na^(+) lt F^(-)`: Increasing ionic sizeB. `B lt C lt N lt O`: Increasing first ionisation enthalpyC. `I lt Br lt F lt CI`: Increasing electron gain enthalpy (magnitude only)D. `Li lt Na lt K lt Rb`: Increasing metallic radius |
| Answer» Correct Answer - A::C::D | |
| 836. |
Pick out the statement(s) which is ( are ) not true about the diagonal relationship of Li and Mg. (i)Polarising powers of `Li^(+)` and `Mg^(2+)` are almost same. (ii) Like Li, Mg decoposes water very fast (iii) LiCl and `MgCl_(2)` are deliquescent. (iv) Like Li,Mg does not form solid bicarbonates.A. Polarising powers of `Li^(+)` and `Mg^(2+)` are almost same .B. Like Li , Mg decomposes water very fastC. LiCl and `MgCl_(2)` are deliquescentD. Like Li, Mg readily reacts with liquid bromine at ordinary temperature . |
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Answer» Correct Answer - B::D |
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| 837. |
In which of the following arrangements, the order is correct according to the property indicated aganst it:a)Increasing size: `Al^(3+) gt Mg^(2+) gt Na^(+) gt F^(-)`b)Increasing `I.E._(1)` : `B lt C lt N gt O`c)Increasing `E.A._(1)` : `I gt Br gt F gt Cl`d)Increasing metallic radius : `Li gt Na gt K gt Rb`A. Increasing size: `Al^(3+) gt Mg^(2+) gt Na^(+) gt F^(-)`B. Increasing `I.E._(1)` : `B lt C lt N gt O`C. Increasing `E.A._(1)` : `I gt Br gt F gt Cl`D. Increasing metallic radius : `Li gt Na gt K gt Rb` |
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Answer» Correct Answer - B |
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| 838. |
In the periodic table, the relation between Li-Mg, Be-Al, B-Si is called diagonal relationship. These pairs of elements show similarities. The factor responsible for this isA. nearly equal electron affinityB. nearly equal polarising powerC. nearly equal electronegativityD. nearly equal ionistion potentials |
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Answer» Correct Answer - B |
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| 839. |
Which of the following factors may be regarded as the main cause of lanthanide contraction?A. Greater shielding of `5d` electrons by `4f` electrons.B. Poorer shielding of `5d` electron by `4f` electrons.C. Effective shielding of one of `4f` electrons by another in the sub-shell.D. Poor shielding of one of `4f` electron by another in the sub-shell |
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Answer» Correct Answer - D Lanthanide contraction is due to poor shielding of one of `4f` electron by another in the sub-shell. |
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| 840. |
Which of the following factors may be regarded as the main cause the lanthanide contraction ?A. Poor shielding of one of 4f electron by another in the subshellB. Effective shielding of one of 4f electrons by another in the subshellC. Poorer shielding of 5d electrons by 4f electronsD. Greater shielding of 5d electrons by 4f electrons |
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Answer» Correct Answer - C [Poorer shielding of 5d electrons by 4f electron] |
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| 841. |
Stability of trivalent and mono valent cation of group 13 (Boron family) will be in order :A. `Ga^(3+) lt In^(3+) lt Tl^(3+)`B. `Ga^(+) gt In^(+) gt Tl^(+)`C. `Ga^(+) lt In^(+) lt Tl^(+)`D. `Ga^(3+) gt Ga^(+)` |
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Answer» Correct Answer - C::D |
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| 842. |
Statement-1: Fluorine forms only one oxoacid, HOF, Statement-2: Fluorine has small size and high electronegativity.A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-17B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-17C. Statement-1 is True, Statement-2 is FalseD. Statement-1 is False, Statement-2 is True |
| Answer» Correct Answer - A | |
| 843. |
Calculate the energy needed to convert three moles of sodium atoms in the gaseous state to sodium ios. The ionization energy of sodium is 495 kJ `mol^(-)`.A. `1485 kJ`B. `495 kJ`C. `148.5 kJ`D. None |
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Answer» Correct Answer - A Energy needed to convert 1 mole of sodium (g) to sodium ions `= 495 kJ` `:.` energy needed to convert 3 mole of `Na(g)` to `Na^(+)` ions `= 495 xx 3 = 1485 kJ` |
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| 844. |
Ionisation energy values of an atom are 495 , 767 , 1250 and 4540 kJ `"mole"^(-1)` the formula of its sulphate isA. `MSO_(4)`B. `M_(2)SO_(4)`C. `M_(2)(SO_(4))_(3)`D. `M(SO_(4))_(2)` |
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Answer» Correct Answer - C |
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