The ${\mathrm{BF}}_3$ is a planar molecule whereas ${\mathrm{NF}}_3$ is pyramidal because:

1. $\mathrm{B}-\mathrm{F}$ bond is more polar than $\mathrm{N}-\mathrm{F}$ bond.

2. Boron atom is bigger than nitrogen atom.

3. Nitrogen is more electronegative than boron.

4. ${\mathrm{BF}}_3$ has no lone pair but ${\mathrm{NF}}_3$ has a lone pair of electrons.

Which of the following statement is true for $I{O}_2{F_2}^{-}$according to VSEPR theory?

(A) The lone pair and two I-O double bonds occupy the equatorial positions of trigonal
bipyramid.

(B) It has $s{p}^3$ hybridization and is T-shaped.

(C)  Its structure is analogous to $S{F}_4$.

(D) (A) and (C) both

The compound with a maximum number of lone pairs of electrons on
the central atom among the following is:

1. $XeO{F}_4$                       

2. $I{F_4}^{+}$

3. $Xe{F}_2$                         

4. $Br{F}_3$

In the structure of $H_{2}CS{F}_4$, to decide the plane in which C = S is present the following bond angle value are given

Axial FSF angle $\left(idealised=180°\right)$            $\Rightarrow 170°$

Equatorial FSF angle $\left(idealised=120°\right)$     $\Rightarrow 97°$

After deciding the plane of double bond, which of the following statements is correct?

(1) Two C-H bonds are in the same plane of axial S-F bonds.

(2) Two C-H bonds are in the same plane of equatorial S-F bonds.

(3) Total five atoms are in the same plane.

(4) Equatorial S-F bonds are perpendicular to plane of $\mathrm{\pi }$ bond.

Which of the following are isoelectronic and isostructural ?

(1) $N{O_2}^{+}, C{O}_2$                                    

(2) $N{O_3}^{-},B{F}_3$

(3) $N{H}_3,C{H_3}^{-}$                                      

(4) All of these

The correct order of acidic strength is :

1. ${\mathrm{Cl}}_2{\mathrm{O}}_7>{\mathrm{SO}}_2>{\mathrm{P}}_4{\mathrm{O}}_{10}$                   

2. ${\mathrm{CO}}_2>{\mathrm{N}}_2{\mathrm{O}}_5>{\mathrm{SO}}_3$

3. ${\mathrm{Na}}_2\mathrm{O}>\mathrm{MgO}>{\mathrm{Al}}_2{\mathrm{O}}_3$                   

4. ${\mathrm{K}}_2\mathrm{O}>\mathrm{CaO}>\mathrm{MgO}$

In which of the following molecules / ions are all the bond angles not equal?

(1) $Si{F}_4$                                                   

(2) $IC{l_4}^{-}$ 

(3) $S{F}_4$                                                    

(4) $PC{l_4}^{+}$

What is the correct order from the weakest to the strongest carbon-oxygen bond for the following species? 

$CO, C{O}_2, C{O_3}^{2-}, C{H}_{3}OH$

(1) $C{H}_{3}OH <C{O_3}^{2-}<C{O}_2<CO$                         

(2) $CO<C{O}_2 <C{O_3}^{2-}<C{H}_{3}OH$

(3) $C{H}_{3}OH <C{O_3}^{2-}<CO<C{O}_2$                         

(4) $C{H}_{3}OH <C{O}_2<C{O_3}^{2-}<CO$

According to Molecular Orbital Theory, which of the following statement is incorrect?

1. C_2​ has no unpaired electrons.    

2. In $C_2$ molecule, both the bonds $\mathrm{\pi }$ are bonds.

3. $C_2$ is paramagnetic but ${C_2}^{2-}$is diamagnetic.     

4. In ${C_2}^{2-}$ ion, there is one $\sigma$ and two $\pi$ bonds.

Match the ionization processes listed in column – I with the changes observed as listed in
column-II. For this use the codes given below:

       Column- I                                                                 Column - II

(a) $N_2 \to N_2^{+}$                                 (p) Bond order increases and magnetic property is changed.

(b) ${O_2}^{+}\to {O_2}^{2+}$                               (q) Bond order decreases and magnetic property is not changed.

(c) $B_2\to {B_2}^{+}$                                   (r) Bond order increases and magnetic property is not changed.

(d) $N{O}^{-}\to NO$                                  (s) Bond order decreases and magnetic property is changed.

Note : Here change in magnetic property refers to change from diamagnetic to paramagnetic
or paramagnetic to diamagnetic. 

                          a      b          c        d
1.                       s       q          p        r
2.                       s       p          q        r
3.                       r       q          s        p
4.                       p       s         q        r