Ncert Exercises & Solutions

Which species lacks tetrahedral geometry?

1. ${BH}_{4}^{-}$

2. ${NH}_{2}^{-}$

3. ${CO}_{3}^{2 -}$

4. $H_{3} O^{+}$

HINT: Molecule having

{sp}^{3}

hybridization has tetrahedral geometry.

Explanation:

STEP 1:

The sum of lone pair and sigma bond represents hybridization. If sum is 4 then hybridization is

{sp}^{3}

, if sum is three than hybridization is

{sp}^{2}

and if sum is two then hybridization is sp.

STEP 2:

{BH}_{4}^{-}

, boron form four sigma bonds and doesnot contains any lone pair . Hence its hybridization is

{sp}^{3}

and geometry is tetrahedral.

{NH}_{2}^{-}

, nitrogen form two sigma bonds and contains two lone pairs . Hence, sum is 4 and hybridization is

{sp}^{3}

. The geometry is tetrahedral.

{CO}_{3}^{2 -}

, carbon form three sigma bonds and doesnot any lone pairs . Hence, sum is 3 and hybridization is

{sp}^{2}

. The geometry is trigonal planar.

H_{3} O^{+}

, oxygen form three sigma bonds and contains one lone pair. Hence, sum is 4 and hybridization is

{sp}^{3}

. The geometry is tetrahedral.

STEP 3:

Hence,

{CO}_{3}^{2 -}

does not has tetrahedral geometry.