
Which among the following is colorless?
A) $S{{c}^{2+}}$
B) $Z{{n}^{2+}}$
c) $T{{i}^{3+}}$
d) ${{V}^{3+}}$
Answer
521.4k+ views
Hint: Transition elements are the elements that are found in Groups 3-12 on the modern periodic table. Transition elements or transition metals have partially filled d or f orbital in any common oxidation state. They are also known as d-block elements.
Complete answer:
The valence electrons of the transition elements (or d- block elements) fall under the d orbital.
General electronic configuration of D-block Elements: $(n-1){{d}^{1-10}}n{{s}^{1-2}}$
The elements which are filling up of $3d,4d,5d,6d$ orbitals:
$3d$- Sc,Ti,V,Cr,Mn,Fe,Co,Ni,Cu,Zn
$4d$-Y,Zr,Nb,Mo,Tc,Ru,Rh,Pd,Ag,Cd
$5d$-La,Hf,Ta,W,Re,Os,Ir,Pt,Au,Hg
$6d$-Incomplete
The Elements mercury, cadmium and zinc are not considered as transition elements because of their electronic configurations $(n-1){{d}^{10}}n{{s}^{2}}$
Colored Transition metal ions have incompletely filled d-orbitals and those that have completely filled d-orbitals or vacant d-orbitals are colorless.
Electronic configuration of Sc and $S{{c}^{2+}}$:
Sc(Atomic No-21) : $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}4{{s}^{2}}$ or $[Ar]3{{d}^{1}}4{{s}^{2}}$
$S{{c}^{2+}}$: $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}$ or $[Ar]3{{d}^{1}}$
Electronic configuration of Zn and $Z{{n}^{2+}}$:
Zn(Atomic No-30): $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}4{{s}^{2}}$or $[Ar]3{{d}^{10}}4{{s}^{2}}$
$Z{{n}^{2+}}$:$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}$ or $[Ar]3{{d}^{10}}$
Electronic configuration of Ti and $T{{i}^{3+}}$:
Ti(Atomic No-22): $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{2}}4{{s}^{2}}$or $[Ar]3{{d}^{2}}4{{s}^{2}}$
$T{{i}^{3+}}$:$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}$or $[Ar]3{{d}^{1}}$
Electronic configuration of V and ${{V}^{3+}}$:
V(Atomic No-23) : $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{3}}4{{s}^{2}}$or $[Ar]3{{d}^{3}}4{{s}^{2}}$
${{V}^{3+}}$:$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{2}}$or $[Ar]3{{d}^{2}}$
As we see that $Z{{n}^{2+}}$ have fully filled d-orbital which means it has no unpaired electrons so it does not undergo electronic transitions. Hence $Z{{n}^{2+}}$ is colorless.
$S{{c}^{2+}}$,$T{{i}^{3+}}$,${{V}^{3+}}$ are colored due to the d-d transition in incompletely filled d-orbitals or it has unpaired electrons which can undergo electronic transitions.
So,the answer is $Z{{n}^{2+}}$ is colorless
Note:
Transition elements typically display metallic qualities such as malleability and ductility. They have high values of electrical conductivity and thermal conductivity and good tensile strength.
When visible light falls on a d-block’s compound or ion, the unpaired electrons of that element which are present in the lower energy d-orbital get excited to higher energy d-orbitals( d-d transition) due to absorption of visible light.
Complete answer:
The valence electrons of the transition elements (or d- block elements) fall under the d orbital.
General electronic configuration of D-block Elements: $(n-1){{d}^{1-10}}n{{s}^{1-2}}$
The elements which are filling up of $3d,4d,5d,6d$ orbitals:
$3d$- Sc,Ti,V,Cr,Mn,Fe,Co,Ni,Cu,Zn
$4d$-Y,Zr,Nb,Mo,Tc,Ru,Rh,Pd,Ag,Cd
$5d$-La,Hf,Ta,W,Re,Os,Ir,Pt,Au,Hg
$6d$-Incomplete
The Elements mercury, cadmium and zinc are not considered as transition elements because of their electronic configurations $(n-1){{d}^{10}}n{{s}^{2}}$
Colored Transition metal ions have incompletely filled d-orbitals and those that have completely filled d-orbitals or vacant d-orbitals are colorless.
Electronic configuration of Sc and $S{{c}^{2+}}$:
Sc(Atomic No-21) : $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}4{{s}^{2}}$ or $[Ar]3{{d}^{1}}4{{s}^{2}}$
$S{{c}^{2+}}$: $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}$ or $[Ar]3{{d}^{1}}$
Electronic configuration of Zn and $Z{{n}^{2+}}$:
Zn(Atomic No-30): $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}4{{s}^{2}}$or $[Ar]3{{d}^{10}}4{{s}^{2}}$
$Z{{n}^{2+}}$:$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{10}}$ or $[Ar]3{{d}^{10}}$
Electronic configuration of Ti and $T{{i}^{3+}}$:
Ti(Atomic No-22): $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{2}}4{{s}^{2}}$or $[Ar]3{{d}^{2}}4{{s}^{2}}$
$T{{i}^{3+}}$:$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{1}}$or $[Ar]3{{d}^{1}}$
Electronic configuration of V and ${{V}^{3+}}$:
V(Atomic No-23) : $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{3}}4{{s}^{2}}$or $[Ar]3{{d}^{3}}4{{s}^{2}}$
${{V}^{3+}}$:$1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}3{{s}^{2}}3{{p}^{6}}3{{d}^{2}}$or $[Ar]3{{d}^{2}}$
As we see that $Z{{n}^{2+}}$ have fully filled d-orbital which means it has no unpaired electrons so it does not undergo electronic transitions. Hence $Z{{n}^{2+}}$ is colorless.
$S{{c}^{2+}}$,$T{{i}^{3+}}$,${{V}^{3+}}$ are colored due to the d-d transition in incompletely filled d-orbitals or it has unpaired electrons which can undergo electronic transitions.
So,the answer is $Z{{n}^{2+}}$ is colorless
Note:
Transition elements typically display metallic qualities such as malleability and ductility. They have high values of electrical conductivity and thermal conductivity and good tensile strength.
When visible light falls on a d-block’s compound or ion, the unpaired electrons of that element which are present in the lower energy d-orbital get excited to higher energy d-orbitals( d-d transition) due to absorption of visible light.
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