MCQ
The ionization energy will be higher when the electron is removed from
- ✓$s-$ orbital
- B$p-$ orbital
- C$d-$ orbital
- D$f-$ orbital
The ionization energy of the elements increases as one moves up a given group because the electrons are held in lower-energy orbitals, closer to the nucleus and therefore are more tightly bound (harder to remove).
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The above reaction is called
$Toluene\,\, \xrightarrow[{2.\,{H^ + }}]{{1.\,KMn{O_4}/O{H^ - }}}I\xrightarrow{{SOC{l_2}}}II\xrightarrow{{N{H_3}}}III\xrightarrow{{OB{r^ - }}}IV$ is
Statement $I$: The boiling point of hydrides of Group $16$ elements follow the order
$\mathrm{H}_2 \mathrm{O}>\mathrm{H}_2 \mathrm{Te}>\mathrm{H}_2 \mathrm{Se}>\mathrm{H}_2 \mathrm{~S}$.
Statement $II$: On the basis of molecular mass, $\mathrm{H}_2 \mathrm{O}$ is expected to have lower boiling point than the othe members of the group but due to the presence of extensive $\mathrm{H}$-bonding in $\mathrm{H}_2 \mathrm{O}$, it has higher boiling point.
In the light of the above statements, choose the correct answer from the options given below:
$(I)$ cis $- [Co(NH_3)_2(en)_2]^{3+}$ $(II)$ trans $-[IrCl_2(C_2O_4)_2]^{3-}$
$(III)\, [Rh(en)_3]^{3+}$ $(IV)$ cis $-[Ir(H_2O)_3Cl_3$