MCQ
Transition metals are not characterized by
- ✓fixed valency
- Bcoloured compound
- Chigh melting and boiling points
- Dtendency to form complexes
✓
Answer
Correct option: A.
fixed valency
a
They do not have fixed valencies since they also have vacant orbitals due to which they show exciting property and when they loose/radiate their energy they impart color in the visible region and also due to availability of vacant orbitals it show wide oxidation numbers and hence make many complex compounds.
They do not have fixed valencies since they also have vacant orbitals due to which they show exciting property and when they loose/radiate their energy they impart color in the visible region and also due to availability of vacant orbitals it show wide oxidation numbers and hence make many complex compounds.
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
Arrange the following compounds in increasing order of basicity$:\ \mathrm{CH}_3 \mathrm{NH}_2,\left(\mathrm{CH}_3\right)_2 \mathrm{NH}, \mathrm{NH}_3, \mathrm{C}_6 \mathrm{H}_5 \mathrm{NH}_2$
→For a $A g-Z n$ button cell, net reaction is $Z n(s)+A g_{2} O(s) \operatorname{Zn} O(s)+2 A g(s)$
→$\Delta G_{f}^{o}\left(A g_{2} O\right)=-11.21\, kJ\,mol ^{-1}$
$\Delta G_{f}^{o}(Z n O)=-318.3\, kJ \,mol ^{-1}$
Then, $E^{o}$ cell of the button cell is.........$V$
$1^o, 2^o, 3^o$ alcohols are distinguished by
→The reaction conditions leading to the best yield of $\mathrm{C}_2 \mathrm{H}_5 \mathrm{Cl}$ are:
→Pure $'N_2'$ obtain by
→Equivalent weight of $H_3PO_2$ when it dispropertinate in to $PH_3$ and $H_3PO_3$ is
→${image}$
→Consider the above chemical reaction and identify product $"A"$
Tertiary butyl alcohol gives tertiary butyl chloride on treatment with
Ozone deplict due to the formation of following compound in Antarctica
Density of aqueous solution of $4\%$ by strength $NaOH$ solution is $1.04\,g/mL$ then molality of solution is
→