Question
Explain the lanthanoid contraction.
✓
Answer
$\rightarrow$ As we move along the lanthanoid series, the atomic number increases gradually by one. This means that the number of electrons and protons present in an atom also increases by one.
$\rightarrow$ As electrons are being added to the same shell, the effective nuclear charge increases.
$\rightarrow$ This happens because the increase in nuclear attraction due to the addition of proton is more pronounced than the increase in the interelectronic repulsions due to the addition of electron.
$\rightarrow$ Also, with the increase in atomic number, the number of electrons in the $4f$ orbital also increases. The $4f$ electrons have poor shielding effect.
$\rightarrow$ Therefore, the effective nuclear charge experienced by the outer electrons increases. Consequently, the attraction of the nucleus for the outermost electrons increases.
$\rightarrow$ This results in a steady decrease in the size of lanthanoids with the increase in the atomic number. This is termed as lanthanoid contraction.
$\rightarrow$ Consequences of lanthanoid contraction
$\rightarrow$ There is similarity in the properties of second and third transition series.
$\rightarrow$ Separation of lanthanoids is possible due to lanthanoid contraction.
$\rightarrow$ It is due to lanthanoid contraction that there is variation in the basic strength of lanthanoid hydroxides. $($Basic strength decreases from $La(OH)_3to Lu(OH)_3.)$
$\rightarrow$ As electrons are being added to the same shell, the effective nuclear charge increases.
$\rightarrow$ This happens because the increase in nuclear attraction due to the addition of proton is more pronounced than the increase in the interelectronic repulsions due to the addition of electron.
$\rightarrow$ Also, with the increase in atomic number, the number of electrons in the $4f$ orbital also increases. The $4f$ electrons have poor shielding effect.
$\rightarrow$ Therefore, the effective nuclear charge experienced by the outer electrons increases. Consequently, the attraction of the nucleus for the outermost electrons increases.
$\rightarrow$ This results in a steady decrease in the size of lanthanoids with the increase in the atomic number. This is termed as lanthanoid contraction.
$\rightarrow$ Consequences of lanthanoid contraction
$\rightarrow$ There is similarity in the properties of second and third transition series.
$\rightarrow$ Separation of lanthanoids is possible due to lanthanoid contraction.
$\rightarrow$ It is due to lanthanoid contraction that there is variation in the basic strength of lanthanoid hydroxides. $($Basic strength decreases from $La(OH)_3to Lu(OH)_3.)$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Write a note on chemical composition of nucleic acid compounds.
For the reaction
$2NO_{(g)} + Cl_{2(g)} \rightarrow 2 NOCl_{(g)}$
The following data were collected. All the measurements were taken at $263K$:
→$2NO_{(g)} + Cl_{2(g)} \rightarrow 2 NOCl_{(g)}$
The following data were collected. All the measurements were taken at $263K$:
| Experiment. No. | Initial $[NO] (M)$ | Initial $[Cl_2] (M)$ | Initial rate of disappearance of $Cl_2$(M/min) |
| $1$ | $0.15$ | $0.15$ | $0.60$ |
| $2$ | $0.15$ | $0.30$ | $1.20$ |
| $3$ | $0.30$ | $0.15$ | $2.40$ |
| $4$ | $0.25$ | $0.25$ | ? |
- Write the expression for rate law.
- Calculate the value of rate constant and specify its units.
- What is the initial rate of disappearance of $Cl_2$ in exp. $4$?
Write the products formed when $(CH_3)_3 C-CHO$ reacts with the following reagents:
$i. \ CH_3COCH_3$ in the presence of dilute $NaOH$
$ii. \text {HCN}$
$iii. NaOH$
→$i. \ CH_3COCH_3$ in the presence of dilute $NaOH$
$ii. \text {HCN}$
$iii. NaOH$
Explain the bromination of Aniline. State what is done if Mono substituted derivative of aniline is to be formed by this process.
Describe the following$:$
$i.$ Acetylation
$ii.$ Cannizzaro reaction
$iii.$ Cross aldol condensation
$iv.$ Decarboxylation
→$i.$ Acetylation
$ii.$ Cannizzaro reaction
$iii.$ Cross aldol condensation
$iv.$ Decarboxylation
A monomer of a polymer upon ozonolysis gives one mole of methylglyoxal and two moles of formaldehyde.
- Identify the monomer of the polymer.
- Give its free radical mode of addition polymerisation.
Explain molecular asymmetry, chirality and enantiomers.
Calculate the boiling point of solution when $4\ g$ of $MgSO_4\ (M = 120\ g\ mol^{-1})$ was dissolved in $100\ g$ of water, assuming $MgSO_4$ undergoes complete ionization.
$(K_b$ for water$=0.52\ K\ kg\ mol^{-1}).$
→$(K_b$ for water$=0.52\ K\ kg\ mol^{-1}).$
What is carbonyl group? Give information about the functional groups containing carbonyl group with their structures.
$a.$ Write the formulation for the galvanic cell in which the reaction $C u(s)+2 A g^{+}(a q) \rightarrow C u^{2+}(a q)+2 A g(s)$ takes place.
Identify the cathode and the anode reactions in it.
$b.$ Write Nernst equation and calculate the emf of the following cell: $Sn ( s )\left| Sn ^{2+}(0.04 M ) \| H ^{+}(0.02 M )\right|$ $H _2(g) \mid \operatorname{Pt}( s )$
$\left(\right.$ Given $\left.E_{S n^{2+} / S n}=-0.14 V\right)$
→Identify the cathode and the anode reactions in it.
$b.$ Write Nernst equation and calculate the emf of the following cell: $Sn ( s )\left| Sn ^{2+}(0.04 M ) \| H ^{+}(0.02 M )\right|$ $H _2(g) \mid \operatorname{Pt}( s )$
$\left(\right.$ Given $\left.E_{S n^{2+} / S n}=-0.14 V\right)$