An element in +2 oxidation state has 24 elect | vedclass

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(B) The electronic configuration of an element with atomic number $Z$ in $+2$ oxidation state is equivalent to an atom with $Z-2$ electrons.Given $Z-2

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$26$ and $4$

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(B) The electronic configuration of an element with atomic number $Z$ in $+2$ oxidation state is equivalent to an atom with $Z-2$ electrons.Given $Z-2 = 24$, the atomic number $Z = 26$. The element is $Fe$.The ground state electronic configuration of $Fe$ $(Z=26)$ is $[Ar] 3d^6 4s^2$.The configuration of $Fe^{2+}$ is $[Ar] 3d^6$.In the $3d^6$ configuration, the electrons are distributed in $5$ orbitals of $d$-subshell according to Hund's rule: $d_{xy}^2, d_{yz}^1, d_{zx}^1, d_{x^2-y^2}^1, d_{z^2}^1$.Thus, there are $4$ unpaired electrons.

An element in $+2$ oxidation state has $24$ electrons. The atomic number of the element and the number of unpaired electrons in it respectively are

Similar Questions

The quantum numbers of four electrons are given below:
$I. \ n = 4, l = 2, m_l = -2, m_s = -1/2$
$II. \ n = 3, l = 2, m_l = 1, m_s = +1/2$
$III. \ n = 4, l = 1, m_l = 0, m_s = +1/2$
$IV. \ n = 3, l = 1, m_l = 1, m_s = -1/2$
The correct order of their increasing energies will be:

An electron in the $5d$ orbital can be represented by the following $(n, l, m_l)$ values:

Chromium has the electronic configuration $4s^1 3d^5$ rather than $4s^2 3d^4$ because

What is the maximum number of orbitals that can be identified with the following quantum numbers?
$n = 3, l = 1, m_l = 0$

The maximum number of electrons in a subshell is given by the expression:

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