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1.	Correct order of 1st ionisationpotential (IP) among following elements Be, B, C, N, O isA. `B lt Be lt C lt O lt N`B. `B lt Be lt C lt N lt O`C. `Be lt B lt C lt N lt O`D. `Be lt B lt C lt O ltN `
Answer» Correct Answer - A An atom has electronic configuration `1s^(2),2s^(2),2p^(6),3s^(2)3p^(6)3d^(3),4s^(2)` It is a member of d- block element because the last electron is filled in d -subshell as `3d^(3)` and the following electronic configuration is possible for d-subshell as `(n-d)^(( 1 "to"10))` `{:(,"Group number",IIIB,IVB,VB,VIB,VIIB),(,,3,4,5,6,7),(,ns^(2)(n-1)s^(2)p^(6),d^(1),d^(2),d^(3),d^(4),d^(5)):}` `{:(,VIII,VIII,VIII,IB,IIB),(,8,9,10,11,12),(,d^(6),d^(7),d^(8),d^(9),d^(10)):}` Hence, it is member of third group

2.	Which of the following oxides is not expected to react with sodium hydroxide ?A. `B_(2)O_(3)`B. `CaO`C. `SiO_(2)`D. `BaO`
Answer» Correct Answer - B Sodium hydroxide, NaOH, being a strong alkali neveneact with a basic oxide (compound). Among the given options, `B_(2)O_(3)` and Beo are amphoteric oxides `, SiO_(2)` is an acidic and `CaO` is a basic oxide. Therefore, NaOH does not react with CaO.

3.	The correct decreasing order of ionic size among the following species is `K^+, Cl^-, S^-2` and `Ca^(+2)`.A. `Ca^(2+) gt K^(+) gt S^(2-) gt Cl^(-)`B. `Cl^(-) gt S^(2-) gt Ca^(2+) gt K^(+)`C. `S^(2-) gt Cl^(-) gt K^(+) gt Ca^(2+)`D. `K^(+) gt Ca^(2+) gt Cl^(-) gt S^(2-)`
Answer» Correct Answer - C Key Idea `"Inico radii" prop "Charge on anion"` `prop (1)/("charge on cation")` During the formation of a cation, the electrons are lost from the outer shell and the remaining electrons experience a great force of attraction by the nucleus, i.e. attracted more towards the nucleus. In other words, nucleus hold the remaining electrons more tightly and this results in decreased radii. However, in case of anion formation, the addition of electron(s) takes place in the same outer shell, thus the hold of nucleus on the electrons of outer shell decreases and this results in increased ionic radii. Thus, the correct order of ionic radii is `S^(2) gt Cl^(-) gt K^(+) gt Ca^(2+)`

4.	Which one of the elements with the following outer orbital configuration may exhibit the larger number of oxidation states ?A. `3d^(3), 4s^(2)`B. `3d^(5), 4s^(1)`C. `3d^(5), 4s^(2)`D. `3d^(2), 4s^(2)`
Answer» Correct Answer - C The sum of number of electrons (unpaired) in d-orbitals and number of electrons ins-orbital gives the number of oxidation states (os) exhibited by a d-block element. Therefore. (a) ` 3d^(3),4s^(2) rArr OS = 3+2=5` (b) `3d^(5),4s^(1)rArr OS = 5+1 = 6` (c) `3d^(5), 4s^(2) rArr OS = 5+ 2=7` (d) `3d^(2), 4s^(2) rArr OS = 2+2=4` Hence, element with `3d^(5), 4s^(2)` configuration exhibits largest number of oxidation states.

5.	The electronic configuration of an element is `1s^(2)2s^(2)2p^(6),3s^(2)3p^(5)`. The atomic number of element present just below the above element in periodic table is:A. 33B. 34C. 36D. 49
Answer» Correct Answer - A The element which present just below the given element will have outermost electromic configuration as `4s^(2) 4p^(3),` so its full electronic configuration is `1s^(2),2s^(2),2p^(6),3s^(2)3p^(6),4s^(2), 3d^(10), 4p^(3)` and hence, its atoimc number is 33.

6.	Which one of the following arrangements represents the correct order of electron gain enthalpy of the given atomic species?A. `Cl lt F lt O lt S`B. `O lt S lt F lt Cl`C. `F lt S lt O lt Cl `D. `S lt O lt Cl lt F`
Answer» Correct Answer - B Key Idea Electron gain enthalpy, generally, increases in a period from left to right and decreases in a group on moving downwards. However, members of III period have somewhat higher electron gain enthalpy as compared to the corresponding members of second period, because of their small size. 0 and S belong to VI A (16) group and Cl and F belong to VII A (17) group. Thus, the electron gam enthalpy of Cl and F is higher as compared to 0 and S. `Cl and F gt 0 and S ` Between Cl and F, Cl has higher electrons gain enthaply then the F, since the incoming electron experience a greater force of repulsion because of small size of F-atom. Similar is true in case of O and S, i.e. the electron gain enthalpy of is higher as compared to O due to its small size. Thus, the correct order of electron gain enthalpy of given elements is `O lt S lt F lt Cl`

7.	`Na^(+), Mg^(2+), Al^(3+)`, and `Si^(4+)` are isoelectronic ions. Their ionic size will follow the orderA. `Na^(+) gt Mg^(2+) lt Al^(3+) lt Si^(4+)`B. `Na^(+) lt Mg^(2+) gt Al^(3+) gt Si^(4+)`C. `Na^(+) gt Mg^(2+) gt Al^(3+) gt Si^(4+)`D. `Na^(+) lt Mg^(2+) lt Al^(3+) ltSi^(4+)`
Answer» Correct Answer - C In isoelectronic species the number of electrons are same but nuclear charge is different. As the nuclear charge increase, the attraction force on last electron increases, so the size decreases or in other words . `"Ionic size" prop(1)/("Charge oncation")`and hence, order is `(Na^(+) gt Mg^(2+) gt Al^(3+) gt Si^(4+))/(("Nuclear charge increase")/("Size decrease"))`

8.	The species `Ar,K^(+)` and `Ca^(2+)` contain the same number of electrons. In which order do their radii increase ?A. `Ar lt K^(+) lt Ca^(2+)`B. `Ca^(2+) lt Ar lt K^(+)`C. `Ca^(2+) lt K^(+) lt Ar`D. `K^(+) lt Ar lt Ca^(2+)`
Answer» Correct Answer - C `Al ^(3+), K^(+)` and `Ca^(2+)` are isoelectronic i.e. with same number of electrons. 18. For isoelectronic species ionic radii decreases with increase in effective (relative) positive charge. Also Ar, K and Ca belong to the same period (3rd period).

9.	The ions `O^(2-), F^(-), Na^(+), Mg^(2+)`, and `A1^(3+)` are isolectronic. Their ionic radii showA. an increase from `O^(2-)` to `F^(-)` and then decrease form `Na^(+)` to `Al^(3+)`B. a decrease form `O^(2-)` to `F^(-)` and then increase form `Na^(+)` to `Al^(3+)`C. a significant increases form `O^(2-)` to `Al^(3+)`D. a significant decrease form `O^(2-)` to `Al^(3+)`
Answer» Correct Answer - D On increasing atomic number of isoelectronic , species ionic radii decreases due to increasing effective nuclear charge `(Z_("eff"))` `"Radius"prop(1)/("Atomic number")prop (1)/(Z_("eff"))` So, as the negative charge increases ionic radii increases while on increasing positive charge ionic radu decreases. Anions having higher ionic radil than the cation `O^(2+) gt F^(-) gt Na^(+) gt Mg^(2+) gt Al^(3+)`

10.	The electronic configuration of four elements are given below. Which element does not belong to the same family as others ?A. `[Xe] 4f^(14), 5d^(10),6s^(2)`B. `[Kr] 4d^(10), 5s^(2)`C. `[Ne] 3s^(2),3p^(5)`D. `[Ar] 3d^(10),4s^(2)`
Answer» Correct Answer - C In a family, all elements have same outermost electronic configuration. Since `[Ne]3s^(2)3p^(5)`, chlorine belongs to halogen family while the remaining three are in same group i.e. group 12. `""_(80)Hg = [Xe]4f^(14) 5d^(10)6s^(2)` `""_(48)Cd = [Kr]4d^(10) 5s^(2)` `""_(30)Zn = [Ar]3d^(10)4s^(2)`

11.	In the periodic table, with the increase in atomic number the metallic nature of elementsA. decrease in a period and increases in a groupB. increases in a period and decreases in a groupC. increase in a period as well as in the groupD.
Answer» Correct Answer - A In periodic table, the metallic character increases down the group because the ionisation enthalpy decreases down the group and metallic character decreases from left to right because the ionisation enthalpy increases from left to right.

12.	The ionisation of hydrogen atom would give rise toA. hydride ionB. hydronium ionC. protonD. hydroxyl ion
Answer» Correct Answer - C Hydrogen have one proton and one electron, when it ionise, i.e. it lose one electron, then only proton is left in the nucleus, so `H^(+)` ion is formed during ionisation which is also called proton. `underset(({:(e^(-)=1),(p=1):}))(H)tounderset("Proton")(H^(+))+e^(-)`

13.	In which of the following options order of arrangement does not agree with the variation of property indicated against it ?A) `Al^(3+) lt Mg^(2+) lt Na^(+) lt F^(-)` (increasing ionic size)B)`B lt C lt N lt O` (increasing first ionisation enthalpy)C)`I lt Br lt Cl lt F` (increasing electron gain enthalpy)D)`Li lt Na lt K lt Rb` (increasing metallic radius)A. `B lt C lt N lt O` (increasing first ionisation enthalpy)B. `I lt Br lt CI lt F` (increasing electron gain enthalpy)C. `Li lt Na lt k lt Rb`( increasing metallic radius )D. `Al^(3+) lt Mg^(2+) lt Na^(+) lt F^(-)` (increasing ionic size)
Answer» Correct Answer - A::B First ionisation energy is the energy required to remove an electron from outermost shell. Hence , correct order is `B lt C lt O lt N` For option (b) Electron gain enthalpy is the energy required to gain an electron in the outermost shell.ltbegt Hence , the correct order is `I lt Br lt F lt CI` For option (c) As we move down the group in alkali metal , metallic radius increases `Li lt Na lt K lt Rb` For option (d) In case of isoelectronic species increases the ionic size of the species increases and vice-versa `Al lt Mg^(2+) lt Na^(+) lt F^(-)`