Question

Which transition metal can form both a high and low spin complex? $Z{n^{2 + }},C{u^{2 + }},M{n^{3 + }},T{i^{2 + }}.$

Answer 1

All that you must know is the concept of the crystal Field Theory of the transition metal and their classification.
You must understand the concept of low spin and high spin and how to fill them both in d orbital.
In this, we can understand these all and by given explanation, you can understand which transition metal can form both a high and low spin complex?

Complete answer:
Manganese metal can form both a high and low spin complex
EXPLANATION: First let us see what type of metal. The crystal Field Theory of the transition metal is typically classified as ${d^1},{d^2},....,{d^{10}}$.
TRANSITION METAL CLASSIFICATIONS
1. $30$ is the atomic number of zinc, so it's on the $10$th column in the transition metals.$\left[ {Ar} \right]4{s^0}3{d^{10}}$ is the electron configuration of $Z{n^{2 + }}$ (just take out the two $4$s electrons)because of this that makes it a ${d^{10}}$ metal.
2. $29$ is the atomic number of copper, so it's on the $9$th column in the transition metals. $\left[ {Ar} \right]4{s^0}3{d^9}$ is the electron configuration of $C{u^{2 + }}$ (just take out the single $4$s electron and the $10$ th $3d$ electron) because of this that makes it a ${d^9}$metals.
3. $25$ is the atomic number of manganese, so it's on the 5th column in the transition metals. $\left[ {Ar} \right]4{s^0}3{d^4}$ is the electron configuration of $M{n^{3 + }}$ (take out the two $4s$ electrons and one $3d$ electron) because of this that makes it a ${d^4}$ metal.
4. $22$ is the atomic number of titanium and so it's on the $2$nd column in the transition metals.$\left[ {Ar} \right]3{d^2}4{s^0}$ is the electron configuration of $T{i^{2 + }}$ (take out the two $4s$ electrons) because of this that makes it a ${d^2}$ metal.
Notice that none of these are ${d^8}$ metals like nickel or platinum, which tend to form square planar or tetrahedral complexes.

Note:
All four of these transition metals commonly have coordination numbers of $6$, however, so let's examine their octahedral complex crystal-field splitting diagrams.
HIGH SPIN and LOW SPIN
High spin = with one electron first fill all five d orbitals and then double up.
Low spin = first fill lowest-energy d orbitals completely and then at last fill higher-energy orbitals.