(a) Phosphorus (P):
The atomic number of phosphorus is 15
Electronic configuration of Phosphorus:
1s 2 2s 2 2p 6 3s 2 3p 3
This can be represented as follows:
From the diagram, it can be observed that phosphorus has three unpaired electrons.
(b) Silicon (Si):
The atomic number of Silicon is 14
Electronic configuration of Silicon:
1s 2 2s 2 2p 6 3s 2 3p 2
This can be represented as follows:
From the diagram, it can be observed that silicon has two unpaired electrons.
(c) Chromium (Cr):
The atomic number of Cr is 24
Electronic configuration of Chromium:
1s 2 2s2 2p 6 3s 2 3p 6 4s 1 3d 5
This can be represented as follows:
From the diagram, it can be observed that chromium has six unpaired electrons.
(d) Iron (Fe):
The atomic number of iron is 26
Electronic configuration of Fe:
1s 2 2s2 2p 6 3s 2 3p 6 4s 2 3d 6
This can be represented as follows:
From the diagram, it can be observed that krypton has no unpaired electrons.
Answered by Pragya Singh | 1 year agoAnswer the following questions:-
(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?
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?
(i) 2s and 3s,
(ii) 4d and 4f,
(iii) 3d and 3p
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?
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 = \( +\dfrac{1}{2}\)