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4 years ago

Write the condensed ground-state electron configurations of these transition metal ions, and state which are paramagnetic:

Chemistry

1 answer:

MAVERICK [17]4 years ago

8 0

Answer and Explanation :

Paramagnetic are those which has unpaired electrons and diamagnetic are those in which all electrons are paired.

(a) V³⁺

The electronic configuration is -

$[Ar]3d^1$

The electrons in 3d orbital = 1 (Unpaired)

Thus, the ion is paramagnetic as the electrons are unpaired.

(b) Cd²⁺

The electronic configuration is -

$[Kr]4d^{10}$

The electrons in 4d orbital = 10 (paired)

Thus, the ion is diamagnetic as the electrons are paired.

The electronic configuration is -

$[Ar]3d^6$

The electrons in 3d orbital = 6 (Unpaired)

Thus, the ion is paramagnetic as the electrons are unpaired.

(d) Ag⁺

The electronic configuration is -

$[Kr]4d^{10}$

The electrons in 4d orbital = 10 (paired)

Thus, the ion is diamagnetic as the electrons are paired.

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A chemist prepares a solution of copper(II) fluoride by measuring out of copper(II) fluoride into a volumetric flask and filling

Simora [160]

The question is incomplete, here is the complete question.

A chemist prepares a solution of copper(II) fluoride by measuring out 0.0498 g of copper(II) fluoride into a 100.0mL volumetric flask and filling the flask to the mark with water.

Calculate the concentration in mol/L of the chemist's copper(II) fluoride solution. Round your answer to 3 significant digits.

Answer: The concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

Explanation:

To calculate the molarity of solute, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

We are given:

Given mass of copper (II) fluoride = 0.0498 g

Molar mass of copper (II) fluoride = 101.54 g/mol

Volume of solution = 100.0 mL

Putting values in above equation, we get:

$\text{Molarity of copper (II) fluoride)=\frac{0.0498\times 1000}{101.54\times 100.0}\\\\\text{Molarity of copper (II) fluoride}=4.90\times 10^{-3}mol/L$

Hence, the concentration of copper fluoride in the solution is $4.90\times 10^{-3}mol/L$

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Explanation:

300c = 1.8f+32= please calculate it yourself ☺

300c= k+273= 573k

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