SolveItClass 9 · NCERT

NCERT Solutions · Class 9 Science Atomic Foundations of Matter

39 questions · 21 still being checked

Revise, Reflect, Refine 9.5–9.6 (part 7 of 22)

  1. Exercise 9.5

    Write the chemical formulae for the following compounds. (i) Aluminium nitrate (ii) Calcium oxide (iii) Ferric oxide
    NCERT’s answer
    (i)
    Al(\(\displaystyle NO_{3}\))$\displaystyle 3$ (ii) CaO (iii) \(\displaystyle Fe_{2}\)\(\displaystyle O_{3}\)
    (i) Aluminium nitrate — \(\displaystyle Al(NO_3)_3\): \(\displaystyle Al^{3+}\) with \(\displaystyle NO_3^-\); criss-cross $\displaystyle 3$ and $\displaystyle 1$ to get one Al to three nitrate ions, and use brackets because more than one polyatomic ion is present.
    (ii) Calcium oxide — CaO: \(\displaystyle Ca^{2+}\) with \(\displaystyle O^{2-}\); criss-crossing gives \(\displaystyle Ca_2O_2\), which is divided by the common factor 2.
    (iii) Ferric oxide — \(\displaystyle Fe_2O_3\): ferric is \(\displaystyle Fe^{3+}\) and oxide is \(\displaystyle O^{2-}\); criss-cross $\displaystyle 3$ and $\displaystyle 2$, with no common factor to remove.

  2. Exercise 9.6

    Write the formulae of the compounds formed from the following pairs of ions. (i) Ca2+\displaystyle Ca^{2+} and Br (ii) Al3+\displaystyle Al^{3+} and CO3\displaystyle CO_{3} (iii) K+\displaystyle K^{+} and SO4\displaystyle SO_{4} (iv) NH4\displaystyle NH_{4} + and Cl
    NCERT’s answer
    (i)
    \(\displaystyle CaBr_{2}\) (ii) \(\displaystyle Al_{2}\)(\(\displaystyle CO_{3}\))$\displaystyle 3$ (iii) \(\displaystyle K_{2}\)\(\displaystyle SO_{4}\) (iv) \(\displaystyle NH_{4}\)Cl
    (i) \(\displaystyle Ca^{2+}\) and \(\displaystyle Br^-\) → \(\displaystyle CaBr_2\) — criss-cross $\displaystyle 2$ and $\displaystyle 1$; \(\displaystyle (2+) + 2 \times (1-) = 0\).
    (ii) \(\displaystyle Al^{3+}\) and \(\displaystyle CO_3^{2-}\) → \(\displaystyle Al_2(CO_3)_3\) — criss-cross $\displaystyle 3$ and $\displaystyle 2$; brackets because three carbonate ions are present.
    (iii) \(\displaystyle K^+\) and \(\displaystyle SO_4^{2-}\) → \(\displaystyle K_2SO_4\) — criss-cross $\displaystyle 1$ and $\displaystyle 2$; only one sulfate ion, so no brackets.
    (iv) \(\displaystyle NH_4^+\) and \(\displaystyle Cl^-\) → \(\displaystyle NH_4Cl\) — both valency $\displaystyle 1$, so one of each and no brackets.