21 + Interview Questions in NATURE-OF-CHEMICAL-BOND IN GENERAL AWARENESS Page 1 InterviewSolution

1.	The correct order of decreasing polarity isA. `HF gt SO_(2) gt H_(2)O gt NH_(3)`B. `HF gt H_(2) O gt SO_(2) gt NH_(3)`C. `HF gt NH_(3) gt SO_(2) gt H_(2)O`D. `H_(2) O gt NH_(3) gt SO_(2) gt HF`
Answer» Correct Answer - B `:.` polarity `prop` difference in electronegativity `:.` Correct order of polarity is `HF gt H_(2)O gt SO_(2) gt NH_(3)`

Discussion

2.	Which of the following set posses `sp^(3)` hybridsatio ?A. `lO_(4)^(-), lCl_(4)^(-), lF_(4)^(+)`B. `XeO_(3), XeO_(4), XeF_(4)`C. `SO_(3)^(2-), SO_(4)^(2-)`D. `PCl_(4)^(+),BF_(4)^(-), lCl_(4)^(-)`
Answer» Correct Answer - C For `SO_(3)^(2-), H=(6+2)/(2)=(8)/(2)=4 (V=6, A=2)` `SO_(4)^(2-) H=(6+2)/(2)4 (V=6, A=2)` `SO_(5)^(2-) H=(6+2)/(2)4 (V=6, A=2)` All have same value of H. So, each on has `sp^(3)` hybridisation.

Discussion

3.	The compound having maximum dipole moment isA. `NH_(3)`B. `NF_(3)`C. `Ncl_(3)`D. `Nl_(3)`
Answer» Correct Answer - A All are isostructural but `NH_(3)` is made up of N and H. Electronegativity difference between N and H is maximum among all, hence `NH_(3)` has highest dipole moment.

Discussion

4.	Which of the following show correct structure of `lCl_(2)`?
Answer» Correct Answer - B To minimies lp-lp- repulsion, they exist as show in option (b) and is stabe in this form of structure.

Discussion

5.	Which is the most covalent?A. `C-F`B. `C-O`C. `C-S`D. `C-Br`
Answer» Correct Answer - B `:.` Covalent character `prop(1)/("lonic character") prop (1)/("Difference in electronegativity")` `:.C-S` is the most convalent.

Discussion

6.	Among the following, molecules, which one have trigonal planar structure ? `XeO_(3), SO_(3), BF_(3), NH_(3)`A. `XeO_(3) and BF_(3)`B. `BF_(2) and SO_(3)`C. `NH_(3) and SO_(3)`D. All of the above
Answer» Correct Answer - B In `BF_(3), H=(3+3)/(2)=3 (sp^(2))` In `SO_(3), H=(6+0)/(2)=3 (sp^(2))`

Discussion

7.	Which se tof molecules are paramagnetic ?A. `B_(2) C_(2) and O_(2)`B. `C_(2), O_(2) and B_(2)`C. `O_(2), N_(2) and B_(2)`D. `B_(2), O_(2) and NO`
Answer» Correct Answer - D MOEC of `B_(2)(10)= sigma 1s^(2), overset()sigma1s^(2), sigma 2s^(2), overset()sigma3s^(2), pi 2p_(x)^(1), ~~ pi 2p_(y)^(1)` MOEC of `O_(2)(16)= sigma1s^(2), overset()sigma 1s^(2), overset()sigma 2s^(2), sigma 2p_(z)^(2)` `pi 2p_(x)^(2) ~~ 2pi_(z)^(2), overset( ) sigma 2p_(x)^(1)~~ overset() sigma 2p_(y)^(1)` MOEC of `NO(15)= sigma 1s^(2), overset() sigma 1s^(2), sigma 2s^(2), overset() sigma 2s^(2), sigma p_(z)^(2)` `pi 2 p_(x)^(2)~~ pi 2p_(y)^(2),overset() pi 2p_(x)^(1)~~ overset() pi 2p_(y)` Due to the presence of unpaired electron, `B_(2), O_(2) and NO` all are paramgnetic.

Discussion

8.	Among the following, chose the correct pair, which is isostructural and isoelectronic ?A. `NO_(3)^(-),CO_(3)^(2-)`B. `SO_(3),NO_(3)^(-)`C. `ClO_(3)^(-),CO_(3)^(2-)`D. `CO_(3)^(2-),ClO_(3)^(-)`
Answer» Correct Answer - A For `NO_(3)^(-)` ion, number of electrons `=7+3xx8+1=32` `H=(V+Y-C+A)/(2)=(5+1)/(2)=3` `rArr` Hybridisation `=sp^(2)` For `CO_(3)^(2-)` ion, number of electrons, `=6+3xx8+2=32` `H=(4+2)/(2)=3` `rArr` Hydrisisation `=sp^(2)` `:.` Both `NO_(3)^(-) and CO_(3)^(2-)` are isoelectronic and isostructural species.

Discussion

9.	Hydrogen bonding is maximum inA. ethyl chlorideB. triethly amineC. ethanolD. diethyl ether
Answer» Correct Answer - C Hydrogen bonding is maximum in ethanol.

Discussion

10.	The d-orbitals involved in `sp^(3)d` hybridisation is:A. `d_(xy)`B. `d_(zx)`C. `d_(z^(2))`D. `d_(x^(2)-y^(2))`
Answer» Correct Answer - B The `D_(zx)` orbtial is involved in `sp^(3)`- hybridisation.

Discussion

11.	In accordance to molecular theory,A. `O_(2)^(+)` is diamagnetic and bond order is more than `O_(2)`B. `O_(2)^(+)` is diamagnetic and bond order is less than `O_(2)`C. `O_(2)^(+)` is diamagnetic and bond order is more than `O_(2)`D. `O_(2)^(+)` is diamagnetic and bond order is less than `O_(2)`
Answer» Correct Answer - C `O_(2)^(+)` contains one unpaired electron and therefore it has bond order of `2.5`while `O_(2)` contains only 2 unpaired electrons. So, it posses bond order of 2.

Discussion

12.	Molecular shapes of `SF_(4). And CF_(4) and XeF_(4)` are:A. different with 1,0 and 2 lone pairs of elcterons on the central atoms, respectivelyB. different with 0,1 and 2 lone pairs of electrons on the central atoms, respectivelyC. the same with 1,1 and 1 lone pairs of electrons on the central atoms, respectivelyD. the same with 2,0 and 1 lone pairs of electrons on the central atoms, respectively
Answer» Correct Answer - A `SF_(4)(sp^(3)d), CF_(4)(Sp^(3)) and (XeF_(4)) (sp^(3)d^(2))` contain 1,0 and 2 lone pairs, respectively. Therefore, their shapes are also different.

Discussion

13.	In `XeF_(2),XeF_(4)` and `XeF_(6)(g)` the number of lone pairs on Xe respectively are :A. 2,3,1B. 1,2,3C. 4,1,2D. 3,2,1
Answer» Correct Answer - D Xe-atom has 8 electrons in its outermost shell. In case of `XeF_(2)`, out of these 8 electrons 2 are used for bond formation, while 3 pairs remains such , i.,e it has 3 lone pair. In case of `XeF_(4),4` electrons of Xe are used for bonding. Therefore, number of lone pairs (non-bonding electrons) is 2. In cae of `XeF_(6),6` electrons are involved for bond formation, thus, number of lone pair is only one.

Discussion

14.	What is the necessary condition for an ionic compound to be soluble in water ?A. `DeltaH_("hyd") gt Delta H_("lattice") `B. `DeltaH_("lattice") gt Delta H_("hyd") `C. `DeltaH_("hyd") gt Delta H_("lattice") `D. None of these
Answer» Correct Answer - B Hydration energy `gt` lattice energy. So that hydration energy minimize the electrostatic repulsion between anion and cation.

Discussion

15.	The correct order of increasing covalent character of the following isA. `SiCl_(4) lt AlCl_(3) lt CaCl_(2) lt KCl`B. `KCl lt CaCl_(2) lt AlCl_(3) lt SiCl_(4)`C. `Al Cl_(3) lt CaCl_(2) lt KCl lt SiCl_(4)`D. None of the above
Answer» Correct Answer - B En differene in `SiCl_(4)=3.0-1.8=1.2` in `AlCl_(3)=3.0-1.5=1.5` in `CaCl_(2)=3.0-1.0=2.0` and in `KCl=3.0-0.8=2.2` `:. KCl lt CaCl_(2) lt AlCl_(3) lt SiCl_(4)` This order is also obtained by applying fajans rule. i.e., caovalent character `prop` charge on ion.

Discussion

16.	The compound `MX_(4)` is tetrahedral. The number of `angleXMX` angles formed in the compound isA. threeB. fourC. fiveD. six
Answer» Correct Answer - D Since, `MX_(4)` is tetrahedral, therefore total number of `angleXMX` is six.

Discussion

17.	Arrange the following ionic compunds in order of increasing ionic character : `{:(,,KF,KCL,KBr,Kl),(,,A,B,C,D):}`A. `A lt B lt C lt D`B. `D lt C lt B lt A`C. `B lt A lt C lt D`D. `C lt A lt B lt D`
Answer» Correct Answer - A In all the given species, capation is common, thus onic character depends upon the size of anion. Smaller the anion, larger is the ionic character of the compound. The order of sizer is `F^(-) lt Cl^(-) lt Br^(-) lt l^(-)`, thus order of ionic character is `KF gt KCl gt KBr gt Kl`.

Discussion

18.	The pair of molecules forming strongest hydrogen bonds areA. `SiH_(4) and SiF_(6)`B. `CH_(3)-underset(O) underset(\|\|)C-CH_(3) and CHCl_(3)`C. `H-underset(O) underset(\|\|) C-OH and CH_(3)- underset(O) underset(\|\|)C-OH`D. `H_(2)O and H_(2)`
Answer» Correct Answer - C Strongest H-bonds are fomed in between `HCOOH and CH_(3)COOH`. This is because H- bonding increase with electronegativity decreases with size or atom.

Discussion

19.	If molecule `MX_3` has zero dipole moment, the sigma bonding orbitals used by M (atomic number `lt21`) areA. sp hybridisedB. `sp^(2)` hybridisedC. `sp^(3)` hybridisedD. None of these
Answer» Correct Answer - B Since, the molecule `(MX_(3))` has zero diple moment, therfore, it must have triangular plnar geometry and accordingly the hybridisation of central metal atom (M) must be `sp^(2)`

Discussion

20.	Bond energy `H-H,F-F and H-F` bonds are 104, 38 and 135 Kcal `"mol"^(-1)`, respectively. The resonance energy in the `H-F` molecule will beA. 142 Kcal `"mol"^(-1)`B. 66 kcal `"mol"^(-1)`C. `72.14` kcal `"mol"^(-1)`D. `79.26` kcal `"mol"^(-1)`
Answer» Correct Answer - C Resonance energy, `Delta _(H-F)=(BE)_(H-F)- sqrt((BE)_(H_(2))(BeE)_(F_(2)))` `=135-sqrt(104xx38)` `=135-62.86=72.14 "kcal mo"^(-1)`

Discussion

21.	The percentage ionic characterin `Cs-Cl` bond present in CsCl molecule will, be, if the electronegativities for Cs and Cl are `0.8 and 3.0`, respectivelyA. `62.9 %`B. `60%`C. `75%`D. `52.14%`
Answer» Correct Answer - D Percentage ionic charcter `=[16(X_(Cl)-X_(Cs))+3.5 (X_(Cl)-X_(Cs))^(2)]` `=[16(3.0-0.8)+3.5(3.0-0.8)^(2)]` `=[16xx2.2+3.5xx(2.2)^(2)]=[35.2+16.94]` `=52.14%`

Discussion

Explore topic-wise InterviewSolutions in .

The correct order of decreasing polarity isA. `HF gt SO_(2) gt H_(2)O gt NH_(3)`B. `HF gt H_(2) O gt SO_(2) gt NH_(3)`C. `HF gt NH_(3) gt SO_(2) gt H_(2)O`D. `H_(2) O gt NH_(3) gt SO_(2) gt HF`

Which of the following set posses `sp^(3)` hybridsatio ?A. `lO_(4)^(-), lCl_(4)^(-), lF_(4)^(+)`B. `XeO_(3), XeO_(4), XeF_(4)`C. `SO_(3)^(2-), SO_(4)^(2-)`D. `PCl_(4)^(+),BF_(4)^(-), lCl_(4)^(-)`

The compound having maximum dipole moment isA. `NH_(3)`B. `NF_(3)`C. `Ncl_(3)`D. `Nl_(3)`

Which of the following show correct structure of `lCl_(2)`?

Which is the most covalent?A. `C-F`B. `C-O`C. `C-S`D. `C-Br`

Among the following, molecules, which one have trigonal planar structure ? `XeO_(3), SO_(3), BF_(3), NH_(3)`A. `XeO_(3) and BF_(3)`B. `BF_(2) and SO_(3)`C. `NH_(3) and SO_(3)`D. All of the above

Which se tof molecules are paramagnetic ?A. `B_(2) C_(2) and O_(2)`B. `C_(2), O_(2) and B_(2)`C. `O_(2), N_(2) and B_(2)`D. `B_(2), O_(2) and NO`

Among the following, chose the correct pair, which is isostructural and isoelectronic ?A. `NO_(3)^(-),CO_(3)^(2-)`B. `SO_(3),NO_(3)^(-)`C. `ClO_(3)^(-),CO_(3)^(2-)`D. `CO_(3)^(2-),ClO_(3)^(-)`

Hydrogen bonding is maximum inA. ethyl chlorideB. triethly amineC. ethanolD. diethyl ether

The d-orbitals involved in `sp^(3)d` hybridisation is:A. `d_(xy)`B. `d_(zx)`C. `d_(z^(2))`D. `d_(x^(2)-y^(2))`

In accordance to molecular theory,A. `O_(2)^(+)` is diamagnetic and bond order is more than `O_(2)`B. `O_(2)^(+)` is diamagnetic and bond order is less than `O_(2)`C. `O_(2)^(+)` is diamagnetic and bond order is more than `O_(2)`D. `O_(2)^(+)` is diamagnetic and bond order is less than `O_(2)`

In `XeF_(2),XeF_(4)` and `XeF_(6)(g)` the number of lone pairs on Xe respectively are :A. 2,3,1B. 1,2,3C. 4,1,2D. 3,2,1

What is the necessary condition for an ionic compound to be soluble in water ?A. `DeltaH_("hyd") gt Delta H_("lattice") `B. `DeltaH_("lattice") gt Delta H_("hyd") `C. `DeltaH_("hyd") gt Delta H_("lattice") `D. None of these

The correct order of increasing covalent character of the following isA. `SiCl_(4) lt AlCl_(3) lt CaCl_(2) lt KCl`B. `KCl lt CaCl_(2) lt AlCl_(3) lt SiCl_(4)`C. `Al Cl_(3) lt CaCl_(2) lt KCl lt SiCl_(4)`D. None of the above

The compound `MX_(4)` is tetrahedral. The number of `angleXMX` angles formed in the compound isA. threeB. fourC. fiveD. six

Arrange the following ionic compunds in order of increasing ionic character : `{:(,,KF,KCL,KBr,Kl),(,,A,B,C,D):}`A. `A lt B lt C lt D`B. `D lt C lt B lt A`C. `B lt A lt C lt D`D. `C lt A lt B lt D`

The pair of molecules forming strongest hydrogen bonds areA. `SiH_(4) and SiF_(6)`B. `CH_(3)-underset(O) underset(||)C-CH_(3) and CHCl_(3)`C. `H-underset(O) underset(||) C-OH and CH_(3)- underset(O) underset(||)C-OH`D. `H_(2)O and H_(2)`

If molecule `MX_3` has zero dipole moment, the sigma bonding orbitals used by M (atomic number `lt21`) areA. sp hybridisedB. `sp^(2)` hybridisedC. `sp^(3)` hybridisedD. None of these

Bond energy `H-H,F-F and H-F` bonds are 104, 38 and 135 Kcal `"mol"^(-1)`, respectively. The resonance energy in the `H-F` molecule will beA. 142 Kcal `"mol"^(-1)`B. 66 kcal `"mol"^(-1)`C. `72.14` kcal `"mol"^(-1)`D. `79.26` kcal `"mol"^(-1)`

The percentage ionic characterin `Cs-Cl` bond present in CsCl molecule will, be, if the electronegativities for Cs and Cl are `0.8 and 3.0`, respectivelyA. `62.9 %`B. `60%`C. `75%`D. `52.14%`