The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?

Asked by Abhisek | 1 year ago |  184

Solution :-

The nuclear charge that is experienced by electrons (which are present in atoms containing multiple electrons) depends on the distance between its orbital and the nucleus of the atom. The greater the distance, the lower the effective nuclear charge. Among p-orbitals, 4p orbitals are the farthest from the nucleus of the bromine atom with (+35) charge. Hence, the electrons that reside in the 4p orbital are the ones to experience the lowest effective nuclear charge. These electrons are also shielded by electrons that are present in the 2p and 3p orbitals along with the s-orbitals.

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

The quantum numbers of six electrons are given below. Arrange them in order of increasing energies.

The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:

1. n = 4, l = 2, $$m_l$$ = –2 , ms = $$-\dfrac{1}{2}$$

2. n = 3, l = 2, $$m_l$$= 1 , ms = $$+\dfrac{1}{2}$$

3. n = 4, l = 1, $$m_l$$ = 0 , ms = $$+\dfrac{1}{2}$$

4. n = 3, l = 2, $$m_l$$= –2 , ms = $$-\dfrac{1}{2}$$

5. n = 3, l = 1, $$m_l$$ = –1 , ms= $$+\dfrac{1}{2}$$

6. n = 4, l = 1, $$m_l$$ = 0 , ms = $$+\dfrac{1}{2}$$