Question

How many d orbitals are involved in hybridization of X?
$A{g_2}S + 2NaCN\mathop \rightleftharpoons \limits^{air} X + N{a_2}S$

A. 3
B. 2
C. 1
D. 0

Answer 1

Hint- In order to solve the question and find the number of d orbitals in the hybridization of X, first of all we will complete the given chemical equation and then on the basis of the compound X. we will check for the positive ion element involved in making bond with negative ion and further on the basis of positive ion we will find the number of d orbitals.

Complete answer:
The given reaction is completed as follows:
$A{g_2}S + 2NaCN\mathop \rightleftharpoons \limits^{air} 2Na\left[ {Ag{{\left( {CN} \right)}_2}} \right] + N{a_2}S$
In the given reaction compound X is $Na\left[ {Ag{{\left( {CN} \right)}_2}} \right]$ .
So the number of d orbitals will be decided on this compound.
In the compound X we can see that silver is involved in a positive ion and makes a bond with negative ion as $A{g^ + }$ in the compound.
So the number of d orbitals will be decided on the basis of $A{g^ + }$ ion.
As we know that $A{g^ + }$ ions hybridization state is $sp$ .
So there are no d orbitals in the compound X.
Hence, 0 d orbitals are involved in hybridization of X.

So, the correct option is D.

Additional information- An orbital is a mathematical term in chemistry and quantum mechanics that describes the wavelike behavior of an electron, electron pair, or (less commonly) nucleon. An orbital can also be called an orbital neutron, or an orbital electron. While most people think of a "ring" with respect to a sphere, the regions of probability density that may hold an electron may be circular, dumbbell-shaped, or more complex three-dimensional shapes.

Note- In chemistry, orbital hybridization (or hybridization) is the process of merging atomic orbitals into new hybrid orbitals (with various energies, sizes, etc.) suitable for combining electrons to form chemical bonds in the theory of valence bonds. Based on the types of orbitals involved in mixing, the hybridization can be classified as $s{p^3},s{p^2},sp,s{p^3}d,s{p^3}{d^2},s{p^3}{d^3}$ . Any central atom surrounded by just two regions of valence electron density in a molecule will exhibit $sp$ hybridization.