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My theory:

The ethyl groups in diethylether being larger in size when compared to hydrogens in water, should be repelled more by the lone pairs on the central atom oxygen. Hence, the ethyl groups should come closer, thereby decreasing the bond angle.

Moreover, the electronegativity difference between the central atom and the peripheral atoms is more in case of water i.e., $\ce{O-H}$ electronegativity difference is more than that in $\ce{C-O}$ as per Pauling’s scale. Hence, the oxygen atom should pull the bond pair closer, thereby allowing to open up the $\ce{H-O-H}$ bond increasing the bond angle.

Hence bond angle of water should be greater than that of diethylether. Am I right?

PS: I know that in reality dimethylether has greater bond angle than water.

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You are mostly correct with your reasoning , and the $\ce{C-O-C}$ bond angle was indeed expected to be lesser than that of $\ce{H-O-H}$ bond angle of water . But it's all theoretical. Experimentally the opposite was observed and was explained on the basis of steric hindrance or steric repulsion of the two bulky $\ce{C2H5}$ groups.

In case of $\ce{(C2H5)2O}$, the positive deviation of the bond angle can be explained on the basis of steric crowding of two ethyl groups. They repel each other and as a result , they do not let the lone pairs of Oxygen compress the bond angle more .

$\ce{H-O-H}$ bond angle is found to be $104.5^\circ$ whereas the $\ce{C-O-C}$ bond angle in diethyl ether is found to be roughly close to $110.5^\circ$. It can be attributed to the steric repulsion that trumps the bond pair-bond pair and lone pair-lone pair repulsion in this case.

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