SolveItClass 9 · NCERT

NCERT Solutions · Class 9 Science Atomic Foundations of Matter

39 questions · 21 still being checked

Pause and Ponder 9.9–9.10 (part 12 of 22)

  1. Exercise 9.9

    The atomic number of fluorine is 9. Explain the formation of the fluorine molecule (F2\displaystyle F_{2}). Just as atoms share electrons to form covalent bonds, we too, can share and care to build strong relationships with people around us. This sharing brings unity and stability, laying the foundation for a stronger community, and ultimately, a strong nation. 170\displaystyle 170 Exploration|Grade 9\displaystyle 9

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    Two fluorine atoms share one pair of electrons, forming a single covalent bond: F—F.
    NCERT_Solution_Class9_Science_Ch9_PP_Q9-9
    Fluorine (\(\displaystyle Z = 9\)) has the configuration $\displaystyle 2$, $\displaystyle 7$ — seven valence electrons, one short of an octet.
    Both atoms need to gain an electron, so neither will donate one; the only way for both to become stable is to share.
    Each atom puts in one electron, and the shared pair is counted by both atoms, so each fluorine now has $\displaystyle 8$ valence electrons.
    The shared pair attracts both nuclei and holds the atoms together; each atom also keeps three lone pairs.
    The molecule is written \(\displaystyle F_2\), or F—F, one line for the one shared pair.

  2. Exercise 9.10

    Show the formation of the following molecules: (i) Carbon dioxide (CO2\displaystyle CO_{2}) (ii) Hydrogen sulfide (H2\displaystyle H_{2}S) (iii) Ammonia (NH3\displaystyle NH_{3})

    Not cross-checked

    NCERT prints no numerical answer for this exercise, so this working has not been cross-checked against the book.

    (i) \(\displaystyle CO_2\) — carbon $\displaystyle (2, 4)$ has $\displaystyle 4$ valence electrons and needs $\displaystyle 4$ more; each oxygen $\displaystyle (2, 6)$ needs 2. Carbon shares two electrons with each oxygen, making two double bonds: O=C=O.
    NCERT_Solution_Class9_Science_Ch9_PP_Q9-10
    (ii) \(\displaystyle H_2S\) — sulfur ($\displaystyle 2$, $\displaystyle 8$, $\displaystyle 6$) needs $\displaystyle 2$ electrons and each hydrogen needs $\displaystyle 1$, so two hydrogen atoms each share one electron with sulfur: two single bonds, H—S—H.
    (iii) \(\displaystyle NH_3\) — nitrogen $\displaystyle (2, 5)$ needs $\displaystyle 3$ electrons, so three hydrogen atoms each share one with it: three single bonds, with one lone pair left on nitrogen.
    Each diagram must show:
    \(\displaystyle CO_2\): C in the middle, an O on each side; $\displaystyle 4$ electrons ($\displaystyle 2$ shared pairs) between C and each O; $\displaystyle 2$ lone pairs left on each O; C ends with $\displaystyle 8$ electrons and each O with $\displaystyle 8$; the line form O=C=O beside it.
    \(\displaystyle H_2S\): S in the middle with an H on each side; $\displaystyle 2$ electrons ($\displaystyle 1$ shared pair) between S and each H; $\displaystyle 2$ lone pairs left on S; each H reaches its duplet of $\displaystyle 2$ and S its octet of $\displaystyle 8$; the line form H—S—H.
    \(\displaystyle NH_3\): N in the middle with three H atoms round it; $\displaystyle 1$ shared pair between N and each H; $\displaystyle 1$ lone pair left on N; N reaches $\displaystyle 8$ and each H reaches 2.
    In all three, use dots for one element's electrons and crosses for the other's, and label every atom with its symbol.