7. On the basis of EAN rule, identify the first-row transition metal for each of

Question

7. On the basis of EAN rule, identify the first-row transition metal for each of the following a. [M(NH3)6]3. Group Number 18-6 x 2NH3 + 3 = 9, so cobalt b. H3CM(CO)s Group Number = 18-5 x 2c0-1 x 2CH3 + 1 [ox] = 7, so manganese c. M(CO)s(PPh3)Br Group Number-18-3 x 2co-1x 2PPh3-1 x 2B-+ 1 [ox) 9, so cobalt d. [M(CN)4 Group Number = 18-4 x 2CN + 2[ox) = 12, so zinc e. [M(CO)4l(en)I Group Number 18-4 x 2co-1 x 2 -1 x 4emt O 4, so titanium [M(CO)o] Group Number = 18-6 x 2co + 1 = 7, so manganese g. [M(CN)4(H20)j" Group Number-18-4 x 2cN-2x 2H20 + 8[ -14, so germanium (not possible) On the basis of the 18-electron rule, determine the expected charge (shown below as "z") and t d electrons on the following a. [Ru(CN)ol 8. b. [PtClol c. [Co(CO)] d. [Ni(CO)4] e. [Mn CO)] f. [Fe(CsHs)2] g. [W(CO)o] d-electron count 18-6x 2CN-6, so Ru(II). z--4 d-electron count = 18-6 x 2c1 = 6, so Pt(IV), z =-2 d-electron count-18-3 x 200-12. z =-3 d-electron count 18-4 x 2co 10, so Ni(dwhy does each Cl atom donate a pair of d-electron count 18-6 x 2co 6, so Mn( How come bromine in 7c and lodine in 7e d-electron count-I 8-2 x 6cp-6, so Fe(II), z 0 d-electron count 18-6 x 2co 6, so w(0)2 = 0 in problem 8b, electrons? donate only one electron.

Explanation / Answer

You are mistaken; see carefully

In 8b, there are 6 chlorines right, each donates 2 electrons

So, it is 12 electrons

Similarly, in 7c. there is only one Br, that donates also 2 electrons

So, it is 2 electrons

Same applies for I in 7e.

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