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QveST [7]
3 years ago
13

Use molecular orbital theory to determine whether f22+ is paramagnetic or diamagnetic.

Chemistry
2 answers:
Alexxandr [17]3 years ago
7 0

We have to know whether F₂²⁺ is paramagnetic or diamagnetic.

F₂²⁺ is paramagnetic.

If number of unpaired electron in any species is equal to zero, the species is diamagnetic and the species contains unpaired electrons, then the species is paramagnetic.

The magnetic property can be explained using molecular orbital theory.

Total number of electron present in   F₂²⁺ is equal to 16 (i.e, 9+9-2). From the molecular orbital electronic configuration, number of electrons present in pi orbitals present is equal to 2.

So,  F₂²⁺ is paramagnetic.

Fynjy0 [20]3 years ago
4 0

Answer : F_2^{2+} is paramagnetic.

Explanation :

According to the molecular orbital theory, the general molecular orbital configuration will be,

(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),[(\pi_{2p_x})=(\pi_{2p_y})],(\sigma_{2p_z}),[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)

As there are 9 electrons present in fluorine.

The number of electrons present in F_2^{2+} molecule = 2(9) - 2 = 16

The molecular orbital configuration of F_2^{2+} molecule will be,

(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,(\sigma_{2p_z})^2,[(\pi_{2p_x})^2=(\pi_{2p_y})^2],[(\pi_{2p_x}^*)^1=(\pi_{2p_y}^*)^1],(\sigma_{2p_z}^*)^0

Paramagnetic compounds : They have unpaired electrons.

Diamagnetic compounds : They have no unpaired electrons that means all are paired.

The number of unpaired electron in molecule is, 2. So, this is paramagnetic. That means, more the number of unpaired electrons, more paramagnetic.

Thus, is paramagnetic.

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The ideal gas heat capacity of nitrogen varies with temperature. It is given by:
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Answer:

A)  1059 J/mol

B)  17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

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C_v=C_p-R

C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314

C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4

A).

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If we integrate above data into the equation; it implies that:

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 1059J/mol

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 17,920 J/mol

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