STD 11 - 4. Chemical bonding and molecular structure — JEE STD 11 Science — Question

$(A)$ The decreasing order of atomic radii of group $13$ elements is $\mathrm{Tl}>\mathrm{In}>\mathrm{Ga}>\mathrm{Al}>\mathrm{B}$.

$(B)$ Down the group $13$ electronegativity decreases from top to bottom.

$(C)$ $\mathrm{Al}$ dissolves in dil. $\mathrm{HCl}$ and liberate $\mathrm{H}_2$ but conc. $\mathrm{HNO}_3$ renders Al passive by forming a protective oxide layer on the surface.

$(D)$ All elements of group 13 exhibits highly stable +1 oxidation state.

$(E)$ Hybridisation of $\mathrm{Al}$ in $\left[\mathrm{Al}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+}$ ion is $\mathrm{sp}^3 \mathrm{~d}^2$.

Choose the correct answer from the options given below:

$(1)$ $Fe^{3+}$ gives brown colour with potassium ferricyanide

$(2)$ $Fe^{2+}$ gives blue colour with potassium ferricyanide

$(3)$ $Fe^{3+}$ gives red colour with potassium thiocyanate

$(4)$ $Fe^{2+}$ gives brown colour with ammonium thiocyanate

$I.$ one atom of oxygen

$II.$ one atom of nitrogen

$III.$ $1\times10^{-10}$ mole of oxygen

$IV.$ $1\times10^{-10}$ mole of copper

$(A)$ The principal quantum number $'n'$ is a positive integer with values of $'n'=1,2,3, \ldots$.

$(B)$ The azimuthal quantum number $'l'$ for a given $'n'$ (principal quantum number) can have values as $'l'=0,1,2, \ldots n$

$(C)$ Magnetic orbital quantum number $'m _{l}'$ for a particular $'l'$ (azimuthal quantum number) has $(2 l$ + 1) values.

$(D)$ $\pm 1 / 2$ are the two possible orientations of electron spin.

$(E)$ For $l=5$, there will be a total of $9$ orbital.

Which of the above statements are correct?

 $HC \equiv CH\mathop {\xrightarrow{{30\%\, {H_2}S{O_4}}}}\limits_{HgS{O_4}} A\xrightarrow{{NaOH}}B$