State the orbital degeneracy of the levels in a hydrogenic atom (Zinparentheses) that have energy (a) −4hcRatom(2); (b) −hcRatom(4), and (c)−hcRatom(5).

(a). 1, (b). 16, and (c). 25.

Explanation:

In quantum Chemistry, When We are are talking about degeneracy as it concerns orbitals we are talking about orbitals that has the same energy.

The energy of the simplest nucleus, which is hydrogen is given below;

Energy (E) = - Z^2 × hcRatom/ n^2. (1).

Also, we know that the orbital degeneracy of the hydrogen atom is = n^2. Therefore, making n^2 the subject of the formula, we have;

n^2 = - Z^2× hcRatom/ Energy (E). (2).

So, let us go!

(a). −4hcRatom(2). Note that we are told that Z is in the parenthesis. Therefore, the degeneracy here is 2;

Degeneracy, n^2 = - Z^2× hcRatom/ Energy (E).

==> - (2)^2 × hc Ratom/ -4hcRatom.

===> - 4 × hcRatom/ - 4 hcRatom.

= 1.

(b). −hcRatom(4), here the Z= 4. So making use of equation (2) again.

Orbital degeneracy= - Z^2× hcRatom/ Energy (E).

===> - (4)^2 ×hcRatom/-hcRarom

= 16.

(c). −hcRatom(5), here, Z= 5. So, making use of equation (2) again.

Orbital degeneracy= - Z^2× hcRatom/ Energy (E).

===> -(5)^2× hcRatom/−hcRatom.

= 25.

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