Explain, giving reasons, which of the following sets of quantum numbers are not possible.

(a) n = 0, l = 0,$$m_l= 0$$$$m_s$$ =$$+\dfrac{1}{2}$$

(b) n = 1, l = 0, $$m_l$$= 0, $$m_s$$$$-\dfrac{1}{2}$$

(c) n = 1, l = 1,$$m_l$$= 0, $$m_s$$ =$$+\dfrac{1}{2}$$

(d) n = 2, l = 0, $$m_l$$= 1, $$m_s$$ =$$-\dfrac{1}{2}$$

(e) n = 3, l = 3, $$ml$$= -3, $$m_s$$ =$$+\dfrac{1}{2}$$

(f) n = 3, l = 0, $$m_l$$= 1, $$m_s$$ =$$+\dfrac{1}{2}$$

Asked by Pragya Singh | 1 year ago |  83

##### Solution :-

(a)  The given set of quantum numbers is not possible because the value of the   principal quantum number (n) cannot be zero.

(b)  The given set of quantum numbers is possible.

(c)  The given set of quantum numbers is not possible. For a given value of n, ‘l’ can have values from zero to (n – 1). For n = 1, l = 0 and not 1.

(d)  The given set of quantum numbers is possible.

(e)  The given set of quantum numbers is not possible. For n = 3,

l = 0 to (3 – 1)

l = 0 to 2 i.e., 0, 1, 2

(f)  The given set of quantum numbers is possible.

Answered by Pragya Singh | 1 year ago

### Related Questions

#### How many sub-shells are associated with n = 4?

(a) How many sub-shells are associated with n = 4?

(b) How many electrons will be present in the sub-shells having ms value of $$- \dfrac{1}{2}$$ for n = 4?

#### Indicate the number of unpaired electrons in:

Indicate the number of unpaired electrons in:

(a) P

(b) Si

(c) Cr

(d) Fe

(e) Kr

#### The unpaired electrons in Al and Si are present in 3p orbital.

The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?

#### Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?

Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge?

(i) 2s and 3s,

(ii) 4d and 4f,

(iii) 3d and 3p