The correct answer is $\boxed{{ – z} \over {z – 5}}$ and the system is unstable.
The system function $H(z)$ of a discrete-time system is defined as the convolution of the impulse response $h[n]$ and the unit step response $u[n]$. In this case, the impulse response is $h[n] = -5nu[- n – 1]$, which is a delayed version of the unit step response with a delay of 1 sample. The unit step response is $u[n] = 1$ for $n \ge 0$ and $u[n] = 0$ for $n < 0$.
The convolution of $h[n]$ and $u[n]$ can be written as
$$H(z) = \sum_{n=-\infty}^{\infty} h[n]u[-n] = \sum_{n=-\infty}^{\infty} -5nu[-n – 1]u[-n] = -5z^{-1}$$
The system is unstable if the poles of $H(z)$ lie outside the unit circle. In this case, the pole of $H(z)$ is at $z = 5$, which lies outside the unit circle. Therefore, the system is unstable.
The other options are incorrect because they do not correspond to the given impulse response.