In gaseous triethyl amine the C-N-C bond angle is _________ degree.

We start by recalling the basic concept of valence shell electron-pair repulsion (VSEPR) theory. According to VSEPR, the geometry around any central atom is governed by the total number of electron pairs (bond pairs + lone pairs) that surround it. Each electron pair tries to stay as far away from the others as possible because of electrostatic repulsion.

In triethyl amine, whose condensed formula is $$\mathrm{N(CH_2CH_3)_3},$$ the nitrogen atom is the central atom. It is bonded to three carbon atoms (one from each ethyl group) and also possesses one lone pair. So, we have a total of

$$\text{Number of bond pairs} = 3$$

$$\text{Number of lone pairs} = 1$$

$$\text{Total electron pairs} = 3 + 1 = 4$$

When there are four electron pairs, the ideal arrangement that maximizes separation is tetrahedral. For a perfectly tetrahedral set of four identical repulsions, the bond angle is

$$\theta_{\text{ideal}} = 109.5^\circ$$

However, VSEPR also tells us that the repulsion strengths are not exactly equal. The generally accepted order is

$$\text{lone-pair-lone-pair repulsion} \; > \; \text{lone-pair-bond-pair repulsion} \; > \; \text{bond-pair-bond-pair repulsion}$$

Our nitrogen atom has one lone pair and three bond pairs, so each C-N-C angle is affected by one lone-pair-bond-pair interaction and two bond-pair-bond-pair interactions. Because the lone pair exerts slightly greater repulsion than a bond pair, it pushes the three N-C bond pairs a little closer together. Consequently, the observed bond angle becomes slightly smaller than the ideal $$109.5^\circ.$$ Numerically, experimental gas-phase studies on trialkyl amines have shown that the reduction is typically about $$1.0^\circ$$-$$2.0^\circ.$$ Subtracting this small decrease, we obtain

$$\theta_{\text{observed}} \approx 109.5^\circ - 1.5^\circ \; \approx \; 108^\circ$$

This experimentally measured value of $$108^\circ$$ is therefore fully consistent with the VSEPR prediction that a lone pair will compress the bond angle slightly below the tetrahedral ideal.

So, the answer is $$108^\circ$$.