score:2

Accepted answer
y.descendants("option").select(z => new option
{
    name = (string)z.attribute("name"),
    value = (string)z.attribute("value")
}

score:0

using (var reader = new stringreader(xml))
{
    var xmlserializer = new xmlserializer(typeof(preferencesmodel));
    var data = xmlserializer.deserialize(reader);
}

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