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Iv) Using dots (.) and Cross (X) Show the bonding in B and Q. v) Explain Why E is a better Conductor of both heat and electricity than B.

Chemistry

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iv) Using dots (.) and Cross (X) Show the bonding in B and Q.

  • Element B: Element B is in Group I (Alkali Metal), meaning it has 1 valence electron.
  • Element Q: Element Q is in Group VIII (Noble Gas), meaning it has 8 valence electrons (a stable octet).

Dot-and-cross diagrams are typically used to show electron sharing in covalent bonds or electron transfer in ionic bonds. For individual elements, they can represent valence electrons.

For Element B:

B⋅\text{B} \cdot

(Element B has one valence electron, which it tends to lose to form a positive ion in chemical reactions, or contributes to a 'sea' of delocalized electrons in metallic bonding.)

For Element Q:

×××Q×××\begin{smallmatrix} \times & \times \\ \times & \text{Q} & \times \\ \times & \times \end{smallmatrix}

(Element Q has eight valence electrons, forming a stable electron configuration. Noble gases like Q are generally unreactive and do not readily form chemical bonds with other elements, nor do they form bonds between their own atoms, existing as discrete atoms.)

v) Explain Why E is a better Conductor of both heat and electricity than B.

Both E (Group III) and B (Group I) are metals, and metals conduct heat and electricity due to the presence of delocalized electrons in their metallic lattice structure.

  • Element E is in Group III, meaning each atom of E contributes three valence electrons to the 'sea' of delocalized electrons.
  • Element B is in Group I, meaning each atom of B contributes only one valence electron to the 'sea' of delocalized electrons.

Since Element E contributes more delocalized electrons per atom than Element B, it has a higher concentration of mobile charge carriers. This greater number of delocalized electrons allows for more efficient transfer of electrical charge and thermal energy, making E a better conductor of both heat and electricity than B.

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