Question

JEE Mains chemistry question is in the image

Answer 1

ANSWER :- -97.

SOLUTION:-

$</p><p> \rightarrow \: [Ti(H_{2} O)_{6}] {}^{ + 3} \\ \\ \\ \implies \: \: Ti {}^{ + 3} = 3d {}^{1} 4s {}^{0} \\ \\ \\ \implies \: t_{2g} {}^{1 \: 0 \: 0}$

$\rightarrow \: CFSE = (-0.4nt_{2g} + 0.6n_{eg}\triangle o + n(p)) \\ \\ \\ \implies \: ( - 0.4 \times 1)20300 = - 8120cm {}^{ - 1} \\ \\ \\ \implies \: \frac{ - 8120}{83.7} kjpmole \: = - 97.$

Hence ,the answer is $\boxed{-97}$

HOPE IT HELPS :)

Answer 2

★ QUESTION :

For [Ti(H2O) 6]³+ the absorption maximum due to d - d transition is found to be 20,300 cm^-1, therefore the Crystal Field stabilization energy in kj/Mole is →

[GIVEN : 1 kJ/mole = 33.7 cm-¹ ]

GIVEN :

• The formula (main) given, that is to be applied = [Ti(H2O) 6]³+

• d - d transition is found which is =

20,300 cm-¹

TO FIND :

• The Crystal Field stabilization energy in Kj/ mole is = ?

STEP-BY-STEP EXPLAINATION :

➠ The formula given = [Ti(H2O) 6]³+

➠ Ti+³ = 3d¹4s^0 (substituting here) (solving)

➠ t2g^100 (Gained value after solving the above)

→ Now, we will have to find here the field crystal (energy) in kJ/mole is →

ㅤㅤㅤㅤㅤas per the given value :

⟹ 1kJ/mole = 33.7 cm-¹

Substituting the value as per the given formula :

⟹ (-0.4×1) 20,300 = 8,120 cm-¹ㅤ

⟹ -8,120/83.7 kJ/mole (when reduced)

ㅤㅤㅤㅤ

⟹ -8,120/83.7 kJ/mole = -97 kJ/mole

Therefore, The crystal field stabilization energy in kJ/mole is = - 97 kJ/mole