91影视

The arrangement of electrons of an atom or molecule is known as the electronic configuration of an atom.

The values of principal quantum number n=1,2,3,.... .The letter used to represent the orbital quantum number I are s,p,d,f,g,... respectively and an orbital q uantum number has n values i.e., 0 to n-1.

If ${\mathbf{Z}}_{\mathbf{eff}}$is the effective atomic number and is the principal quantum number, the energy level of an electron is:

${\mathbf{E}}_{\mathbf{n}}\mathbf{=}\mathbf{-}\frac{{\mathbf{Z}}_{\mathbf{eff}}^{\mathbf{2}}}{{\mathbf{n}}^{\mathbf{2}}}\left(13.6\mathrm{eV}\right)$鈥︹赌︹赌︹赌︹赌�.(1)

The shell holds electrons,p the shell holds 6 electrons and d the shell holds the 10 electrons.

(a)

The Nitrogen atom has 7 electrons. To find the number of electrons in ${\mathrm{N}}^{2+}$, remove 2 electrons from nitrogen. So,${\mathrm{N}}^{2+}$ has 5 electrons, and the ground-state electronic configuration of${\mathrm{N}}^{2+}$ is:

$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^1$

(b)

The effective atomic number${\mathrm{Z}}_{\mathrm{eff}}=7-4=3$ as the nucleus has the charge 7+e and there are 4 inner electrons with charge -4e. The principal quantum number n=2 for the 2p state.

Substitute the values in the equation (1):

$$\begin{gathered} \mathrm{E}=-\frac{{\left(3.0\right)}^2}{{\left(2\right)}^2}\left(13.6\mathrm{eV}\right) \\ =-30.6\mathrm{eV} \end{gathered}$$

Hence, the ground-state electronic configuration of ${\mathrm{N}}^{2+}$is$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^1$ and the energy of the least strongly bound level in the L shell of${\mathrm{N}}^{2+}$ is -30.6 eV.

(c)

The phosphorus atom has electrons. To find the number of electrons in ${\mathrm{P}}^{2+}$, remove electrons from phosphorus. So, ${\mathrm{P}}^{2+}$has 13 electrons, and the ground-state electronic configuration of ${\mathrm{P}}^{2+}$is:

$1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^1$

(d)

The effective atomic number ${\mathrm{Z}}_{\mathrm{eff}}=15-12=3$ as the nucleus has the charge +15e and there are 12 inner electrons with charge -12e. The principal quantum number n=3 for the 3p state.

Substitute the values in the equation (1):

$$\begin{gathered} \mathrm{E}=-\frac{{\left(3.0\right)}^2}{{\left(3\right)}^2}\left(13.6\mathrm{eV}\right) \\ =-13.6\mathrm{eV} \end{gathered}$$

Hence, the ground-state electronic configuration of ${\mathrm{P}}^{2+}$is $1{\mathrm{s}}^{2}2{\mathrm{s}}^{2}2{\mathrm{p}}^{6}3{\mathrm{s}}^{2}3{\mathrm{p}}^1$and the energy of the least strongly bound level in the M shell of${\mathrm{P}}^{2+}$ is -13.6 eV.