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The provided table shows the first 11 elements of a new periodic system. Identify the elements based on their group and period. Specifically, answer: a) Which two elements are in the same group and ha

Chemistry

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2 parts
d(i)

BH4\text{BH}_4

e(iii)
10.5

Step-by-step solution

2 steps

Part a

Elements in the same group have similar chemical properties.

From the table, C and F are both in Group I.
Therefore, elements with similar chemical properties are C and F.

Part b

Transition metals are located in the d-block of the periodic table, typically between Group II and Group III (or Groups 3-12).

Based on its position spanning across the middle of the third period, G is a transition metal.

Part c

Elements that hardly form compounds are noble gases, which are found in Group VIII.

From the table, E is in Group VIII.

Part d(i)

BH4\text{BH}_4

Element B (Group IV) has 4 valence electrons and forms 4 bonds. Hydrogen has 1 valence electron and forms 1 bond.

Element B is in Group IV. Hydrogen is in Group I.

The formula of the compound is BH4\boxed{\text{BH}_4}.

Part d(ii)

B is a non-metal and hydrogen is a non-metal. The bond formed between two non-metals is a *covalent bond

Part d(iii)

Dot and cross diagram for BH4\text{BH}_4

B is the central atom with 4 valence electrons (represented by dots). Each H atom has 1 valence electron (represented by crosses). Each H forms a single covalent bond with B.

H×H⋅B×⋅H⋅H\begin{array}{c} \text{H} \\ \times \\ \text{H} \quad \cdot \underset{\times}{\text{B}} \cdot \quad \text{H} \\ \quad \quad \cdot \\ \quad \quad \text{H} \end{array}

More accurately, showing shared pairs:

H×H×B⋅⋅×H⋅H\begin{array}{c} \text{H} \\ \times \\ \text{H} \quad \times \underset{\cdot}{\text{B}} \cdot \times \quad \text{H} \\ \quad \quad \cdot \\ \quad \quad \text{H} \end{array}

Each bond is a shared pair of electrons, one from B and one from H.

H×⋅H×⋅B⋅×H×⋅H\begin{array}{c} \text{H} \\ \times \cdot \\ \text{H} \quad \times \cdot \text{B} \cdot \times \quad \text{H} \\ \quad \quad \times \cdot \\ \quad \quad \text{H} \end{array}

Part e(i)

Atoms of the same element with different mass numbers are called *isotopes

Element D occurs naturally as D10\text{D}^{10} and D12\text{D}^{12} in the ratio 75 : 25 respectively.

Part e(ii)

Isotopes differ in the number of neutrons. Therefore, the neutron is the sub-atomic particle responsible for this phenomenon.

Part e(iii)10.5

To determine the relative atomic mass of D, we use the weighted average of the masses of its isotopes

  1. Convert percentages to decimal abundances.

    Abundance of D10=75%=0.75\text{D}^{10} = 75\% = 0.75
    Abundance of D12=25%=0.25\text{D}^{12} = 25\% = 0.25

  2. Calculate the relative atomic mass.

    Relative atomic mass=(mass of D10×abundance of D10)+(mass of D12×abundance of D12)\text{Relative atomic mass} = (\text{mass of } \text{D}^{10} \times \text{abundance of } \text{D}^{10}) + (\text{mass of } \text{D}^{12} \times \text{abundance of } \text{D}^{12}) Relative atomic mass=(10×0.75)+(12×0.25)\text{Relative atomic mass} = (10 \times 0.75) + (12 \times 0.25) Relative atomic mass=7.5+3.0\text{Relative atomic mass} = 7.5 + 3.0 Relative atomic mass=10.5\text{Relative atomic mass} = \boxed{10.5}
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