The increasing order of basicity for the following intermediates is (from weak to strong):

1. Electronegativity and Hybridization:

   As the s-character of a carbon atom increases ($$\text{sp}^3 = 25\%$$, $$\text{sp}^2 = 33.3\%$$, $$\text{sp} = 50\%$$), its electronegativity increases. A more electronegative atom holds its lone pair of electrons more tightly, making it more stable and less willing to donate them to a proton (less basic). Therefore, basicity follows the order:

   $$\text{sp}^3\text{-hybridized C}^- > \text{sp}^2\text{-hybridized C}^- > \text{sp-hybridized C}^-$$

2. Inductive Effect ($$+I$$):

   Alkyl groups are electron-donating groups via the $$+I$$ effect. In $$\text{sp}^3$$ carbanions, the presence of electron-donating methyl groups increases electron density on the negative carbon, increasing its tendency to donate electrons and making it highly basic.

Detailed Stability and Basicity Breakdown:

• $$(v)\text{ }\text{CN}^-$$ (Cyanide ion): The negative charge resides on a highly electronegative nitrogen atom, and the carbon is $$\text{sp}$$-hybridized. This is exceptionally stable, making it the weakest base.
• $$(iii)\text{ }\text{HC}\equiv\text{C}^-$$ (Acetylide ion): The negative charge is on an $$\text{sp}$$-hybridized carbon ($$50\%$$ s-character). It holds the charge tightly, making it a weak base.
• $$(ii)\text{ }\text{CH}_2=\text{CH}^-$$ (Vinyl carbanion): The negative charge is on an $$\text{sp}^2$$-hybridized carbon ($$33.3\%$$ s-character). It is more basic than $$\text{sp}$$ but less basic than $$\text{sp}^3$$.
• $$(iv)\text{ }\text{CH}_3^-$$ (Methyl carbanion): An $$\text{sp}^3$$-hybridized carbanion ($$25\%$$ s-character) with no additional alkyl groups destabilizing it. It is strongly basic.
• $$(i)\text{ }(\text{CH}_3)_3\text{C}^-$$ (tert-Butyl carbanion): An $$\text{sp}^3$$-hybridized carbanion flanked by three electron-donating methyl groups ($$+I$$ effect). This drastically destabilizes the negative charge, making it the most unstable intermediate and the strongest base.

Conclusion:

Combining these factors gives the following increasing order of basicity (weak to strong):

$$\mathbf{(v) < (iii) < (ii) < (iv) < (i)}$$

Answer: Option C