Pair of transition metal ions having the same number of unpaired electrons is:

Solution & Explanation

### Core Logic Let's map the electronic configurations and count the unpaired d-orbital electrons for each option: * For pair (1): V^2+ implies [Ar] 3d^3 4s^0 implies 3 text unpaired electrons Co^2+ implies [Ar] 3d^7 4s^0 implies t_2g^5 e_g^2 implies 3 text unpaired electrons Both ions contain exactly 3 unpaired electrons. * For other ions: Ti^2+ implies [Ar] 3d^2 implies 2 text unpaired e-, quad Fe^3+ implies [Ar] 3d^5 implies 5 text unpaired e- Cr^2+ implies [Ar] 3d^4 implies 4 text unpaired e-, quad Ti^3+ implies [Ar] 3d^1 implies 1 text unpaired e- Mn^2+ implies [Ar] 3d^5 implies 5 text unpaired e- ### Pattern Recognition D-orbital counts follow a predictable symmetry: a 3d^n system contains the same number of unpaired electrons as a 3d^10-n system under high-spin conditions. This explains why 3d^3 (V^2+) and 3d^7 (Co^2+) match perfectly with 3 unpaired electrons each. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements

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The D And F Block Elements

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Q86 jee_main_2024_31_jan_evening Chromyl Chloride Test
In the reaction of potassium dichromate, potassium chloride and sulfuric acid (conc.), the oxidation state of the chromium in the product is (+) ________
Numerical Answer. Answer: 6 to 6

Solution

### Related Formula K_2Cr_2O_7(s) + 4KCl(s) + 6H_2SO_4(conc.) rightarrow 2CrO_2Cl_2(g) + 6KHSO_4 + 3H_2O ### Core Logic This reaction represents the Chromyl Chloride test used to detect the presence of chloride ions. When potassium dichromate is heated with a metal chloride in concentrated sulfuric acid, red vapors of chromyl chloride (CrO_2Cl_2) are evolved. ### Step 1: Oxidation State Calculation In chromyl chloride (CrO_2Cl_2): Let the oxidation state of Chromium be x. Oxygen is typically -2 and Chlorine is -1. x + 2(-2) + 2(-1) = 0 x - 4 - 2 = 0 x = +6 Thus, the oxidation state of Chromium in the product is 6. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements Class 11 Chemistry: Practical Chemistry
Q71 jee_main_2024_31_jan_morning Potassium Dichromate and Permanganate
Identify correct statements from below: A. The chromate ion is square planar. B. Dichromates are generally prepared from chromates. C. The green manganate ion is diamagnetic. D. Dark green coloured K_2MnO_4 disproportionates in a neutral or acidic medium to give permanganate. E. With increasing oxidation number of transition metal, ionic character of the oxides decreases. Choose the correct answer from the options given below:
  • A. textB, C, D only
  • B. textA, D, E only
  • C. textA, B, C only
  • D. textB, D, E only

Solution

### Step 1: Statement A Analysis CrO_4^2- (chromate ion) is tetrahedral, not square planar. Statement A is incorrect. ### Step 2: Statement B Analysis 2Na_2CrO_4 + 2H^+ rightarrow Na_2Cr_2O_7 + 2Na^+ + H_2O. Dichromates are indeed prepared from chromates. Statement B is correct. ### Step 3: Statement C Analysis The green manganate ion (MnO_4^2-) has manganese in the +6 oxidation state (3d^1). Thus, it contains 1 unpaired electron and is paramagnetic, not diamagnetic. Statement C is incorrect. ### Step 4: Statement D Analysis Dark green coloured K_2MnO_4 undergoes disproportionation in neutral or acidic media to yield permanganate (MnO_4^-) and manganese dioxide (MnO_2). Statement D is correct. ### Step 5: Statement E Analysis Fajans' rule dictates that as the oxidation state increases, polarizing power increases, leading to a decrease in ionic character (increase in covalent character). Statement E is correct. ### Final Conclusion The correct statements are B, D, and E. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Chemistry: The d- and f-Block Elements

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