Beryllium has 2 valence electrons. Which element could pair with it in a 1 Be : 1 atom ratio? A P(5valenceelectrons) B. Se(6valenceelectrons) C Li(1valenceelectron) D. Br(7valenceelectrons)

Beryllium (Be) has 2 valence electrons and typically forms a \( \text{Be}^{2+} \) ion by losing these two electrons. To pair with beryllium in a \( 1:1 \) atom ratio, we need an element that can accept 2 electrons to achieve a stable electron configuration.<br /><br />Let's evaluate the options:<br /><br />A. Phosphorus (P) has 5 valence electrons and needs 3 more electrons to complete its octet, so it is not suitable for pairing with Be in a \( 1:1 \) ratio.<br /><br />B. Selenium (Se) has 6 valence electrons and needs 2 more electrons to complete its octet. This makes Se a suitable candidate to pair with Be in a \( 1:1 \) ratio, as it can accept the 2 electrons that Be loses.<br /><br />C. Lithium (Li) has 1 valence electron and tends to lose this electron to form a \( \text{Li}^+ \) ion. It cannot pair with Be in a \( 1:1 \) ratio because it does not have the capacity to accept 2 electrons.<br /><br />D. Bromine (Br) has 7 valence electrons and needs only 1 more electron to complete its octet. Therefore, it would pair with Be in a \( 1:2 \) ratio, not \( 1:1 \).<br /><br />The correct answer is B. Selenium (Se), which can pair with beryllium in a \( 1 \mathrm{Be}: 1 \) atom ratio.