Occurrence,characteristics of Transition Elements

Why Titanium and Vanadium passive towards dilute non oxidising acids?

Platinum (IV) compounds are thermodyanamically more stable than nickel (IV) compounds.

Explain thermodynamic stability of transition metal compounds on the basis of ionization enthalpy.

Explain about the thermodynamic stability of transition element.

Give reason for the following-

Nickel (II) compounds are thermodynamically more stable than platinum (II) compounds.

Give reason for the following-

Platinum (IV) compounds are more stable than nickel (IV) compounds.

Explain the occurrence of more electropositive hard metals of d-block.

Explain the occurrence of copper metal of d-block.

Explain the occurrence of titanium metal of d-block.

Explain the occurrence of vanadium metal of d-block.

Explain the occurrence of soft metals of d-block.

Why transition elements are good conductors of heat and electricity?

Transition elements are good conductors of heat and electricity.

Find the metallic and ferromagnetic substance.

Predict the correct order of the metallic character among the following:

Which element has the highest first ionisation potential?

Give a reason why $\mathrm{Au}\left(79\right)$, $\mathrm{Ag}\left(47\right)$ have almost the same Ionization potential.

The chloride $\mathrm{‘}\mathrm{X}\mathrm{’}$ of a metal gives a black precipitate with ${\mathrm{NH}}_4\mathrm{OH}.$ The chloride $\mathrm{‘}\mathrm{Y}\mathrm{’}$ of same metal gives a white precipitate with ${\mathrm{NH}}_4\mathrm{OH}.$ With $\mathrm{KI}$, $\text{'}\mathrm{Y}\text{'}$ gives red precipitate which is soluble in excess of $\mathrm{KI}.$ The formula of chlorides $\mathrm{‘}\mathrm{X}\mathrm{’}$ and $\mathrm{‘}\mathrm{Y}\mathrm{’}$ respectively are:

On which of the following factors, stability of oxidation state of a cation in aqueous medium depends? 

Choose the correct statement(s) among the following.

(i) The third ionization enthalpy of $\mathrm{Mn}$$(\mathrm{Z}=25)$ is exceptionally low.

(ii)  ${\mathrm{Cr}}^{2+}$ is a stronger reducing agent than ${\mathrm{Fe}}^{2+}$.