Read the two statements below about the compounds H2Y, Li2Y, HZ and
KZ and then answer the questions that follow:
A non-metal element which lies in the third row of the Periodic Table
forms a gaseous compound H2Y with hydrogen and the compound
Li2Y with lithium, which has a solid crystalline structure and a high
melting point (greater than 900 °C).
• A non-metal element which lies in the fourth row of the Periodic Table
forms a gaseous compound HZ with hydrogen and the compound KZ
with potassium, which has a solid crystalline structure and a high
melting point (greater than 700 °C)
(i)Identify to which Group of the Periodic Table the elements Y and Z
belong and hence write down the name and chemical symbol of Y and
Z. Give brief reasoning on how you arrived at your answers, including
the likely valency of Y and Z in the compounds given.
(ii) What type of bonding is involved in compounds Li2Y and KZ? Give
your reasoning.
(iii) State two different conditions under which you would expect
compounds Li2Y and KZ to conduct electricity. Explain your answer
in terms of the type of bonding found in the compounds.
I think Y is S and Z is Br. Now you write all of the equations. Check my thinking.
To answer these questions, we need to analyze the given information and make use of our knowledge about the Periodic Table and chemical bonding.
(i) Based on the information given, we can determine the Group of the Periodic Table to which the elements Y and Z belong.
Y: According to the information, the non-metal element Y lies in the third row of the Periodic Table. Elements in the same row have the same number of electron shells. Since Y forms a compound with hydrogen (H2Y), it is likely to be in Group 17, also known as the Halogens. The chemical symbol for a halogen in Group 17 is X, so Y must be an element in Group 17.
Z: Similarly, the non-metal element Z lies in the fourth row of the Periodic Table. Again, elements in the same row have the same number of electron shells. Z forms a compound with hydrogen (HZ) and potassium (KZ). Elements in Group 14 of the Periodic Table have 4 valence electrons, making them a likely candidate. The chemical symbol for an element in Group 14 is C, so Z must be an element in Group 14.
Therefore, the name and chemical symbols are:
Y – Chlorine (Cl)
Z – Carbon (C)
(ii) To determine the type of bonding involved in compounds Li2Y and KZ, we need to consider the elements involved and their likely valency.
Li2Y: Lithium (Li) is an alkali metal with a valence of +1, and Y is a halogen with a valence of -1. When an alkali metal combines with a halogen, they typically form an ionic bond due to the transfer of electrons. So, Li2Y likely has ionic bonding.
KZ: Potassium (K) is also an alkali metal with a valence of +1, and Z is carbon (C) from Group 14. Carbon typically forms covalent bonds with other non-metals. Since KZ involves an alkali metal and carbon, it is likely to have ionic bonding.
(iii) The ability of a compound to conduct electricity depends on the type of bonding present.
Li2Y: Ionic compounds can conduct electricity when dissolved in water or in a molten state. In the case of Li2Y, it has an ionic bond between Li+ and Y-. Hence, under conditions where Li2Y is dissolved in water or melted, it would conduct electricity.
KZ: Covalent compounds, like KZ, do not conduct electricity in a solid state because the electrons are held tightly within the covalent bonds. However, if KZ were to be melted or dissolved in a suitable solvent, it might form ions and conduct electricity.
To summarize:
(i) Y belongs to Group 17 and is chlorine (Cl). Z belongs to Group 14 and is carbon (C).
(ii) Li2Y has ionic bonding, while KZ has ionic bonding as well.
(iii) Li2Y can conduct electricity when dissolved in water or melted, while KZ may conduct electricity when melted or dissolved in a suitable solvent.