i need to deduce the type of bonding that holds sodium cyanide together
and write a balanced equation for sodium cyanide dissolving in water
any help would be appreciated
sarah
Na^+ + :C:::N:^-
Na to CN bond is ionic (more or less).
C-N bond is covalent.
NaCN + HOH ==> NaOH + HCN.
The Na is not involved. The net ionic equation is
CN^- + HOH ==> HCN + OH^-
thank you very much for getting back to me but i'm more confused now
how do you get the balanced chemical equation NaOH + HCN
i also need to state whether it is covelent or ionic the text i have worked from doesn't say which is involved in these bonds it just wants one or the other
i'm beginning to think i should have chosen a different subject.
I have the same question. I took write a balanced equation for the reaction that occurs when sodium cyanide dissolves in water as a separate question, to the type of bonding.
Read the beginning of the question about the chemicals that will give you some clues about answering it.
The electronegativity of C is 2.55 on a chart I have and the EN of Na is 0.9. The difference is 1.65 which is VERY close to a 50% covalent-50% ionic bond. Technically we would call this a covalent bond. USUALLY when a question asks about a bond in a molecule the bond between the cation and the anion is the one the focus on. Since I don't know your text or your prof's views, I don't know which answer to give. Polar covalent would be my best answer if we ignore the bond for the anion which is the CN^- ion. And the break even point for covalent/ionic character is also murcky; i.e., some use 1.7, some 1.9, and some 2.1. To add confusion, some call the bond ionic if it is more than 50% ionic and they call it covalent if it is more than 50% covalent and don't quibble about polar covalent. Your job is to take all of this information in and answer according to your notes and your text's views. As for the equation, NaCN is the salt of a strong base (NaOH) and a weak acid (HCN). When NaCN is placed in water, it hydrolyzes (reacts with) water. Actually, the Na^+ doesn't react; only the CN^- does. The net ionic equation for the hydrolysis of CN^- is
CN^- + HOH ==> HCN + OH^-.
CN^- is a stronger Bronsted-Lowry base than water; therefore, it pulls a H^+ away from HOH to form HCN. The OH^- remains. That means that a solution of NaCN is basic BECAUSE OH^- are produced. You can even calculate the ionization constant for CN^- as a base. It is Kw for water/Ka for HCN.
To summarize:
The type of bonding in sodium cyanide (NaCN) involves both ionic and covalent bonds. The bond between Na+ and CN- is predominantly ionic, as sodium (Na) is a metal and cyanide (CN) is a polyatomic ion with a negative charge. The bond between carbon (C) and nitrogen (N) in the cyanide ion is covalent.
The balanced equation for sodium cyanide dissolving in water is:
NaCN + HOH -> NaOH + HCN
In this reaction, NaCN hydrolyzes in water, resulting in the formation of sodium hydroxide (NaOH) and hydrogen cyanide (HCN). It's important to note that the Na+ does not react in this equation, only the CN- ion.
As for whether the bonding in NaCN is more ionic or more covalent, it can be challenging to determine definitively without specific information from your text or professor. The electronegativity difference between Na and C is around 1.65, indicating a bond that is close to being equally covalent and ionic. Depending on your specific criteria or definitions, you can argue for either a predominantly covalent or predominantly ionic bond.
Overall, it's important to consider the properties of the elements involved, electronegativity differences, and the characteristics of the resulting compounds when determining the type of bonding in a molecule like sodium cyanide.