SATHEE: Chemical Bonding and Molecular Structure

Chemical Bonding and Molecular Structure - Result Question 73

####75. The $A s F_{5}$ molecule is trigonal bipyramidal. The hybrid orbitals used by the As atom for bonding are

(a) $d_{x^{2} - y^{2}}, d_{z^{2}}, s, p_{x}, p_{y}$

[1997]

(b) $d_{x y}, s, p_{x}, p_{y}, p_{z}$

(d) $d_{x^{2} - y^{2}}, s, p_{x}, p_{y}, p_{z}$

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Solution:

$A s = 1 s^{2} 2 s^{2} 2 p^{6} \underset{↓ s p^{3} d hybridisation}{\underset{⏟}{3 s^{1} 3 p_{x}^{1} 3 p_{y}^{1} 3 p_{z}^{1} 3 d^{1}}}$

So, the hybrid orbitals used by As atom in $A s F_{5}$ molecule are $s, p_{x}, p_{y}, p_{z}, d_{z}^{2}$ .

In the formation of a stable trigonal pyramid, the two axial bonds above and below the equatorial xy-plane have to be equally strong. Thus, the $d_{z}^{2}$ orbital is used.

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Chemical Bonding and Molecular Structure - Result Question 73

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