In $P{O}_4^{-3}$, formal charge on every oxygen atom and P-O bond order respectively are:
1.  -0.75 and 1.25
2.  -0.5 and 2
3.  1 and 1.5
4.  -0.75 and 2

Among the following, the pair in which the two species are not isostructural is:

1. ${\mathrm{IO}}_3^{-}\text{and }{\mathrm{XeO}}_3$

2. ${\mathrm{BH}}_4^{-}\text{and }{\mathrm{NH}}_4^{+}$

3. ${\mathrm{PF}}_6^{-}\text{and }{\mathrm{SF}}_6$

4. ${\mathrm{SiF}}_4\text{ and }{\mathrm{SF}}_4$

Increasing order of bond length is:

1. $N{O}^{-}$ $<$ $NO$ $<$ $N{O}^{+}$ $<$ $O_2^{-}$

2. $O_2^{-}$ $<$ $NO$ $<$ $N{O}^{-}$ $<$ $N{O}^{+}$ $$

3. $O_2^{-}$ $<$ $N{O}^{-}$ $<$ $NO<$ $N{O}^{+}$ $$

4. $N{O}^{+}$ $<$ $NO$ $<$ $N{O}^{-}$ $<$ $O_2^{-}$

H₂O is dipolar, whereas BeF₂ is not. It is because:
 

In a regular octahedral molecule, MX₆, the number of X–M–X bonds at 180º is:

The lone pairs in the BrF₃ molecule, occupy equatorial positions to minimize:

The dipole moment of compound AB is 10.92 D and that of compound CD is 12.45 D. If the bond length of AB is 2.72 A° and that of CD is 2.56 A°, then for these compounds the correct statement is:
1.  More ionic nature in AB
2.  More ionic nature in CD
3.  Equal in both
4.  Cannot be predicted

Among the following, iso-electronic species are:

1. CO₂, NO₂

2. $N{O}_2^{-}$, CO₂

3. CN^–, CO

4. SO₂, CO₂

 ${\mathrm{p}}_{\mathrm{\pi }}-{\mathrm{d}}_{\mathrm{\pi }}$ bonding is present in:

1. ${\mathrm{NO}}_3^{-}$

2. ${{\mathrm{SO}}_3}^{-2}$

3. ${{\mathrm{BO}}_3}^{-3}$

4. ${{\mathrm{CO}}_3}^{-2}$

In NO₃^– ion, the number of bond pair and lone pair of electrons on nitrogen atom are respectively: