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Anastaziya [24]
3 years ago
10

For the following give (1) oxidation # of metal, (2) number of d electrons, draw valence bond description of the complex, fill i

n metal and ligand valence electrons, give (3) metal orbitals that are hybridized, (4) type of hybridization, (5) molecular geometry, and (6) draw the crystal field de- scription of the octahedral complexes for a), c), and e).
[Cr(H2O)6]2+ (High spin)
(1) ___ (2) ____ (3) ______ (4) ________ (5) ___________

Chemistry
1 answer:
andrezito [222]3 years ago
4 0

Answer:

1) the oxidation state of the metal is +2

2) there are four d electrons

3) metal valence electrons =4 ligand valence electrons= 18

4) sp^3d^2

5) octahedral geometry

Explanation:

[Cr(H2O)6]2+ is an octahedral complex. Octahedral complexes are known to have the metal ion in sp^3d^2 hybridization state. Since there are six ligands each bonding to the central metal ion, there are eighteen ligand valence electrons and four valence electrons from the metal present in the high spin complex.

The crystal field of high spin and low spin octahedral for a d4 ion is shown in the image attached.

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5.712 grams = kilograms
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0.005712

Explanation:

5.712 grams = 0.005712 kilograms

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3 years ago
When the concentration of a reactant is increased, the
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3 0
3 years ago
Calculate the EMF between copper and silver Ag+e-E=0.89v<br>Cu=E=0.34v​
bogdanovich [222]

Answer:

Depending on the E^\circ value of \rm Ag^{+} + e^{-} \to Ag\; (s), the cell potential would be:

  • 0.55\; \rm V, using data from this particular question; or
  • approximately 0.46\; \rm V, using data from the CRC handbooks.

Explanation:

In this galvanic cell, the following two reactions are going on:

  • The conversion between \rm Ag\; (s) and \rm Ag^{+} ions, \rm Ag^{+} + e^{-} \rightleftharpoons Ag\; (s), and
  • The conversion between \rm Cu\; (s) and \rm Cu^{2+} ions, \rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s).

Note that the standard reduction potential of \rm Ag^{+} ions to \rm Ag\; (s) is higher than that of \rm Cu^{2+} ions to \rm Cu\; (s). Alternatively, consider the fact that in the metal activity series, copper is more reactive than silver. Either way, the reaction is this cell will be spontaneous (and will generate a positive EMF) only if \rm Ag^{+} ions are reduced while \rm Cu\; (s) is oxidized.

Therefore:

  • The reduction reaction at the cathode will be: \rm Ag^{+} + e^{-} \to Ag\; (s). The standard cell potential of this reaction (according to this question) is E(\text{cathode}) = 0.89\; \rm V. According to the 2012 CRC handbook, that value will be approximately 0.79\; \rm V.
  • The oxidation at the anode will be: \rm Cu\; (s) \to \rm Cu^{2+} + 2\, e^{-}. According to this question, this reaction in the opposite direction (\rm Cu^{2+}\; (aq) + 2\, e^{-} \rightleftharpoons \rm Cu\; (s)) has an electrode potential of 0.34\; \rm V. When that reaction is inverted, the electrode potential will also be inverted. Therefore, E(\text{anode}) = -0.34\; \rm V.

The cell potential is the sum of the electrode potentials at the cathode and at the anode:

\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &= 0.89 \; \rm V + (-0.34\; \rm V) = 0.55\; \rm V\end{aligned}.

Using data from the 1985 and 2012 CRC Handbook:

\begin{aligned}E(\text{cell}) &= E(\text{cathode}) + E(\text{anode}) \\ &\approx 0.7996 \; \rm V + (-0.337\; \rm V) \approx 0.46\; \rm V\end{aligned}.

5 0
4 years ago
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