Question

how many valence electrons does the following element have? 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p³

Explanation:

Step1: Identify the valence shell

The electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^3\). The valence shell is the outermost shell, which is the 4th shell here (since \(n = 4\) for \(4s\) and \(4p\)).

Step2: Sum valence electrons

In the valence shell (n = 4), we have \(4s^2\) and \(4p^3\). The number of valence electrons is the sum of electrons in these subshells. So, \(2 + 3 = 5\)? Wait, no, wait. Wait, the electron configuration: let's check again. Wait, the electron configuration is \(1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^{10} 4p^3\). Wait, no, actually, the correct way is that for main group elements, valence electrons are in the outermost \(s\) and \(p\) subshells. Wait, the element here: let's find the highest principal quantum number \(n\). Here, \(n = 4\) (from \(4s\) and \(4p\)). The electrons in \(4s^2\) and \(4p^3\): \(2 + 3 = 5\)? But the options have 3, 15, 5, 13. Wait, maybe I made a mistake. Wait, no, wait the electron configuration: let's count the total electrons first. \(2 + 2 + 6 + 2 + 6 + 2 + 10 + 3 = 33\). The element with atomic number 33 is arsenic (As), which is in group 15? Wait, no, group 15 has 5 valence electrons? Wait, no, group number for main group elements: for \(p\)-block, group number is \(10 +\) number of \(p\) electrons. Wait, \(4p^3\), so \(10 + 3 = 13\)? No, that's not right. Wait, no, the valence electrons for elements in the \(p\)-block: the number of valence electrons is equal to the group number. Wait, arsenic is in group 15? Wait, no, atomic number 33: electron configuration is \([Ar] 3d^{10} 4s^2 4p^3\). So the valence shell is \(4s^2 4p^3\), so total valence electrons are \(2 + 3 = 5\)? But the options have 5 as option C? Wait, the options are: A. 3, B. 15, C. 5, D. 13. So the correct answer should be 5, because the valence electrons are in the outermost shell (n=4), which has \(4s^2\) and \(4p^3\), so 2 + 3 = 5.

Answer:

C. 5