Organometallic Compounds - Chemistry MCQ

At high temperature(30⁰C) due to fast 1,2-meta shift,  mode looks like  mode. Hence, two signal. At low temperature, this rate is very slow, hence four signal corresponding to 4-types of hydrogen atom. Hence, (X and Y).

Bridging carbonyl is more basic than terminal hence Lewis acid attack at bridging carbonyl. Hence, correct option is (a)

In B due to presence of methyl group, it will donate more electron density to metal. Hence, more back donation therefore, less value of v _c-o. 

For all Cp to similar environment all the CO should be in μ³- mode

As C₂F₄ is strong -acceptor than C₂H₄. Hence, there will be more backbonding in (Y). Hence, C–C bond will be larger and M–C bond will be shorter in Y.

At –118ºC movement of axial ligand to equatorial is slow. Hence, axial and equatorial ligands give different signal. Hence, intensity ratio will be 2 : 3

[Re₂Cl₄(PMe₂Ph)₄]⁺
Let the oxidation states of two Re atoms are x and y
x + y - 4 = + 1
x + y = +5 
x = +2, y = +3  

Therefore, total number of d-electrons on Re²⁺ and Re3* ions = 5 + 4 = 9 Electronic configuration in this cluster ion is

Bond order 

This reactions reductive elimination takes place first then followed by oxidative addition.

In above reaction CH₂ get inserted  between two metal. Hence, reaction is an example of insertion reaction.

Complex A obey 18 electron, in Co(Cp)₂ there is an electron in antibonding orbital which leads to increase in bond length. Hence, correct option is (2) and (3).

Grubbs catalyst used for ring closing metathesis of olefins.

Each transition metal has 12 electrons associated with it that are reserved for non-framework bonding. Electron more than 12 electrons are involved in cluster bonding.
Number of valence electron in Co(CO)₄  =  17
Hence, number of electrons towards cluster bonding = 17–12 = 5

Let the oxidation state of Pd be ��. The oxidation state of the (OAc) group is -1. The overall compound is neutral.

Hence, �� + 2(-1) = 0

�� = 2 = oxidation state of Pd.