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Ann [662]
3 years ago
6

2. Among the complexes K4[Cr(CN)6] (A),K4[Fe(CN)6] (B) , K3[Co(CN)6] (C) , K4[Mn(CN)6] (D), Jahn-Teller distortion is shown by *

D (0/1 Points) A,B and C B,C and D A and D B and C
Chemistry
1 answer:
Vanyuwa [196]3 years ago
6 0

Answer:

K4[Cr(CN)6] and K4[Mn(CN)6]

Explanation:

The Jahn–Teller effect, sometimes also known as Jahn–Teller distortion, describes the geometrical distortion of molecules and ions that is associated with certain electron configurations. In Jahn Teller effect, more electrons occupy the lower orbital than the upper one, resulting in an overall lowering of the electronic energy. This distortion is typically observed among octahedral complexes where the two axial bonds can be shorter or longer than those of the equatorial bonds. In octahedral complexes, the Jahn–Teller effect is most pronounced when an odd number of electrons occupy the eg orbitals. This situation is most pronounced in complexes with the configurations d9 , low-spin d7 or high-spin d4. These configurations display a strong Jahn Teller effect. Jahn Teller may also be observed if there is a degeneracy due to the electrons in the t2g orbitals. A weak Jahn Teller effect is also observed for d5 species.

K4[Cr(CN)6] contains Cr^2+ which is a d4 specie, hence it exhibits Jahn Teller effect. Also, K4[Mn(CN)6] contains Mn^2+ which is a d5 specie, therefore it also exhibits Jahn Teller effect.

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3 years ago
(part 1 of 3) Copper reacts with silver nitrate through a single replacement. If 1.29 g of silver are produced from the reaction
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Answer:

See explanation.

Explanation:

Hello there!

In this case, according to the described chemical reaction, we first write the corresponding equation to obtain:

Cu+2AgNO_3\rightarrow 2Ag+Cu(NO_3)_2

Thus, we proceed as follows:

Part 1 of 3: here, since the molar mass of silver and copper (II) nitrate are 107.87 and 187.55 g/mol respectively, and the mole ratio of the former to the latter is 2:1, we can set up the following stoichiometric expression:

m_{Cu(NO_3)_2}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu(NO_3)_2}{2molAg}*\frac{187.55gCu(NO_3)_2}{1molCu(NO_3)_2}   \\\\m_{Cu(NO_3)_2}=1.12gCu(NO_3)_2

Part 2 of 3: here, the molar mass of copper is 63.55 g/mol and the mole ratio of silver to copper is 2:1, the mass of the former that was used to start the reaction was:

m_{Cu}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{1molCu}{2molAg}*\frac{63.55gCu)_2}{1molCu}   \\\\m_{Cu}=0.380gCu

Part 3 of 3: here, the molar mass of silver nitrate is 169.87 g/mol and their mole ratio 2:2, thus, the mass of initial silver nitrate is:

m_{AgNO_3}=1.29gAg*\frac{1molAg}{107.87gAg}*\frac{2molAgNO_3}{2molAg}*\frac{169.87gAgNO_3}{1molAgNO_3}   \\\\m_{AgNO_3}=2.03gAgNO_3

Best regards!

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3 years ago
To what volume should you dilute 50.0 ml of 12 m hno3 solution to obtain a 0.100 m hno3 solution?
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Answer:

The answer is "6L"

Explanation:

Formula:

\bold{C_1 \times V_1 = C_2 \times V_2 }\\\\V_2= \frac{C_1\times V_1}{C_2}

Values:

\to C_1= 12 \ m\\\to V_1= 50 \ ml\\\to C_2= 0.100 \ m\\\\\\V_2= \frac{12 \times 50 }{0.100}

   = \frac{12 \times 50 }{0.100}\\\\= \frac{12 \times 50 \times 1000}{100}\\\\= \frac{600 \times 1000}{100}\\\\= 600 \times 10\\\\=6000 \ ml\\= 6 \ L

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